Measure the resistance of the phase zero loop. Methods for measuring the impedance of a phase-zero circuit. Live measurement

The article discusses the method for calculating the resistance of a phase-zero circuit in electrical installations with voltages up to 1000 V with a solidly grounded neutral and the rules for calculating the short circuit current in the line, which allows you to check the agreement of the circuit parameters with the characteristics of protection devices in electrical installations. The data presented in the article is intended primarily for calculations of distribution and group networks.

To perform calculations of short circuit currents in transformer substations, it is necessary to additionally take into account the type, power, connection diagram, and voltage at the transformer input. Therefore, the use of this work to calculate transformer substations will only allow approximately estimating their parameters.

In general, the resistance of the circuit phase zero R L - N is equal to:

where Z t /3 is the resistance of the transformer, Ohm; R Σ per - total contact resistance, Ohm; R Σ aut - total resistance of all circuit breakers, Ohm; R n - resistivity of the nth section of the circuit Ohm/km (according to Table 1); L n - length of the nth section of the chain, km; R arc - arc resistance at the short circuit, Ohm.

Table 1

Cross-section of phase conductors mm 2	Cross section of the neutral core mm 2	Phase circuit impedance - zero, Ohm/km at cable core temperature of +65 degrees
		Core material:
		Aluminum
				Z chain (cable)			Z chain (cable)

table 2

Transformer power, kV∙A
Transformer resistance, Zt/3, Ohm (Δ/Υ)

Table 3

I no. auto off, A										50 or more

Table 4

R circuit, Ohm

When designing a group network, if the supply and distribution networks have already been laid, it is advisable to measure the resistance of the phase-zero circuit from the transformer to the buses of the group panel. This can significantly reduce the likelihood of errors in group network calculations. In this case, the resistance is calculated using the formula:

R L - N = R dist + R per. gr + R auto gr + Rn gr ∙Ln gr + Rarc (2)

where, R dis - measured resistance of the phase - zero circuit of the line connected to the input circuit breaker of the group panel, Ohm; R AC - resistance of transition contacts in the group line, Ohm; R aut.gr - total resistance of circuit breakers - input group panel and outgoing group line, Ohm; Rn gr - cable resistivity n-th group lines (according to table 1), Ohm/km; Ln gr - length of the nth group line, km.

Let's consider the process of calculating the resistance of the phase - zero circuit of the circuit shown in Fig. 1 with a single-phase short circuit phase to zero at the end of the group line.

Initial data:

A transformer with a power of 630 kV∙A is connected according to the “delta-star” circuit - according to Table 2 we find Zt/3=0.014 Ohm;

Supply network - a cable with aluminum conductors 80 meters long has a phase conductor of 150 mm 2 and a neutral conductor of 50 mm 2. From Table 1 we find the cable resistivity to be 0.986 Ohm/km. We calculate its resistance (we express the cable lengths in kilometers): 0.986 Ohm/km∙0.08 km=0.079 Ohm;

Distribution network - cable with copper cores 50 meters long and core cross-section 35 mm 2. From Table 1 we find the cable resistivity to be 1.25 Ohm/km. We calculate its resistance:

1.25 Ohm/km∙0.05 km=0.0625 Ohm;

Group network - a cable with copper conductors 35 meters long and a conductor cross-section of 2.5 mm 2. From Table 1 we find the cable resistivity to be 17.46 Ohm/km. We calculate its resistance:

17.46 Ohm/km∙0.035 km=0.61 Ohm;

The outgoing line circuit breaker is 16 Amperes (with response characteristic “C”), the input circuit breaker of the group panel is 32 Amperes, the remaining circuit breakers in the line have a rated current of more than 50 Amps. We calculate their resistance (according to Table 3) 0.01 Ohm+0.004 Ohm+3∙0.001 Ohm=0.017 Ohm;

We will take into account the transition resistance of the contacts only in the group line (the connection points of the group line cable to the panel and to the load). We get 2∙0.01 Ohm=0.02 Ohm.

We sum up all the obtained values ​​and get the resistance of the phase - zero circuit without taking into account the arc resistance R L - N = 0.014 + 0.079 + 0.0625 + 0.61 + 0.017 + 0.02 = 0.80 Ohm.

From Table 4 we take the arc resistance of 0.075 Ohm, and we obtain the final value of the desired value R L - N =0.80 Ohm+0.075 Ohm=0.875 Ohm.

The Electrical Installation Rules (PUE) specify the longest circuit shutdown time during a short circuit in networks with a solidly grounded neutral: 0.2 seconds at a voltage of 380 V and 0.4 seconds at a voltage of 220 V.

To ensure the specified response time of the protection, it is necessary that in the event of a short circuit in the protected line, a current arises that exceeds at least 3 times the rated current of the fuse link of the nearest fuse (for explosive premises, at least 4 times) and at least 3 times current of the circuit breaker release, which has a characteristic inversely dependent on the current (for explosive premises no less than 6 times). For circuit breakers with a combined release (having a thermal release for protection against overloads and an electromagnetic release for protection against short-circuit currents), the short-circuit current must exceed the operating current of the electromagnetic release by at least 1.2 - 1.25 times.

Currently, automatic circuit breakers are used with different current ratios between the electromagnetic release and the thermal release. Automatic switches of group “B” have a multiplicity ranging from 3 to 5, group “C” from 5 to 10, group “D” from 10 to 20, group “K” from 10 to 15 and group “Z” from 2 to 3 . When calculating, it is always taken maximum value multiplicity of trip currents of releases. For example, for circuit breaker C16, the short circuit current must be at least 16 A∙10∙1.2=192 A (for circuit breaker C10 at least 10A∙10∙1.2=120 A and for C25 at least 25 A∙10∙ 1.2=300 A). In the above example, we obtained a phase-to-zero circuit resistance of 0.875 Ohms. With such a circuit resistance, the short circuit current Is will be equal to

U f / R L - N = 220V/0.875 Ohm = 251 A. Therefore, the group line in the given example is protected from short circuit currents.

The maximum resistance of the phase-zero circuit for the C16 circuit breaker will be 220 V/192A=1.14 Ohm. In the given example of a network (Fig. 1), the resistance of the circuit from the transformer to the buses of the group panel will be 0.875 Ohm - 0.61 Ohm = 0.265 Ohm. Therefore, the maximum possible resistance of the group line cable will be 1.14 Ohm - 0.265 Ohm = 0.875 Ohm. Its maximum length L with a cable cross-section of 2.5 mm 2 will be determined using Table 1.

L, km=0.875 Ohm/(17.46 Ohm/km)=0.050 km.

Whenever possible, you should calculate the group network with a maximum margin for the resistance of the phase-zero circuit, especially the socket network. Often loads (iron, kettle and others Appliances), in which short circuits often occur, are connected to the outlet via an extension cord. Starting from a certain length of the extension cord, the coordination of the circuit parameters with the characteristics of the protection devices is disrupted, that is, it turns out to be insufficient to instantly shut down the network. The emergency section can be switched off only by a thermal release after a relatively long period of time (several seconds), as a result of which the cables can heat up to an unacceptably high temperature. high temperatures until the insulation ignites.

The electrical wiring design must be carried out in such a way that even if the cable insulation ignites due to a short circuit, this does not lead to a fire. That is why they began to lay hidden electrical wiring in steel pipes in buildings with building structures made from flammable materials. In explosive buildings, it is advisable to use more complex cable protection against exposure.

In accordance with PTEEP, to control the sensitivity of protection to single-phase ground faults in installations up to 1000 V with a solidly grounded neutral, it is necessary to measure the resistance of the phase-zero loop.

To measure the resistance of the “phase-zero” loop, there are a number of devices that differ in circuits, accuracy, etc. The areas of application of various devices are given in Table. 1.

Measuring instruments electrical parameters grounding devices, including measuring the resistance of the phase-zero loop

The check is carried out for the most remote and most powerful electrical receivers, but not less than 10% of their total number.

The check can be done by calculation using the formula Zpet = Zp + Zt / 3 where Zp is the total resistance of the wires of the phase-zero loop; Zt is the total resistance of the supply transformer. For aluminum and copper wires Zpet = 0.6 Ohm/km.

The current of a single-phase short circuit to earth is determined by Zpet: Iк = Uф / Zpet If the calculation shows that the multiplicity of the current of a single-phase ground fault is 30% higher than the permissible multiplicity of operation of protective devices specified in , then we can limit ourselves to calculation. IN otherwise direct measurements of the short-circuit current should be carried out using special devices, for example, types EKO-200, EKZ-01 or using the ammeter-voltmeter method at reduced voltage.

The ammeter-voltmeter method for measuring the resistance of a phase-zero loop

The electrical equipment under test is disconnected from the network. The measurement is made on alternating current from a step-down transformer. To measure, an artificial short circuit of one phase wire is made to the housing of the electrical receiver. The test diagram is shown in the figure.

After applying voltage, the current I is measuredand voltage U, measuring current must be at least 10 - 20 A. Resistance of the measured loop Zп=U/I. The resulting value of Zp must be arithmetically added with the calculated value of the impedance of one phase of the supply transformer R t/3.

Program for measuring phase-zero loop resistance

1. Familiarization with design and as-built documentation and the results of previous tests and measurements.

2. Preparation of the necessary electrical measuring instruments and testing devices, conductors and protective equipment.

3. After completing organizational and technical measures and access to the facility, performing measurements and tests

4. Evaluation and processing of measurement and test results.

5. Registration of measurements and tests.

6. Correction of diagrams, execution of signatures on the suitability (not suitability) of electrical equipment for further operation.

Briefly, phase-zero loop resistance measured to determine the behavior of circuit breakers when a short circuit occurs. A short circuit occurs when the cable is mechanically damaged or the cable insulation is destroyed due to aging. In electrical installations with a grounded neutral, the neutral conductor is connected to the neutral of the transformer. The transformer neutral is connected to the ground loop. When a phase is shorted to a phase, to a body or to a zero, an electrical circuit is created. This circuit is called a phase-zero loop. With an interphase fault, the current in the circuit will be greater than with a single-phase fault. The resistance of the phase-zero loop should be as small as possible, then the short circuit current in the loop will be greatest and the protection will operate faster. Phase-zero loop measurement and currents of short phase circuits is carried out to determine the response time protective devices. Based on the obtained value of the resistance of the phase-zero loop, the short-circuit current value is obtained by calculation. The operating time of the protection device depends on the current value. The protection device is usually a circuit breaker. The response time of the machine must meet the requirements of the electrical installation rules. If this time does not go beyond 5 seconds for 380 Volts and 0.4 seconds for 220 Volts, then roughly the line protection can be considered sufficient. Automatic power off should provide protection against damage electric shock in case of indirect touches and short circuits. The faster the circuit breaker operates, the less damage will be caused to people and wiring in the electrical installation, because during a short circuit the current value instantly increases and the temperature of the conductor increases sharply. At the same time, the insulation begins to melt and burn. Even a few seconds of idle protection may be enough to damage and ignite tens of meters of cable, since a damaged cable ignites neighboring cables. IN Lately During installation, a non-flammable cable is used, which helps prevent fires, but does not save the wiring from damage or the premises from smoke. If desired, you can use a low-smoke cable, but financial conditions do not always allow this. The resistance of the phase-zero loop is affected by the length of the line, the cross-section of the line conductors, the method of connecting sections of the line, the quality of the line laying, and the number of bolted connections. Along with checking the protection devices themselves phase-zero measurement gives good result in ensuring the safety of electrical installations.

© All materials are protected by the Russian Federation copyright law and the Civil Code of the Russian Federation. Full copying is prohibited without permission from the resource administration. Partial copying is permitted with a direct link to the source. Author of the article: team of engineers from OJSC Energetik LTD

Electricity supply

Checking the operating conditions of the protective device during a single-phase fault in networks with voltages up to 1000V with solid grounding of the neutral

IN electrical networks voltage up to 1000 V with solid grounding of the neutral must ensure reliable shutdown of a single-phase short circuit by a protective device. This is dictated by safety requirements.
Calculation points for determining the magnitude of the short-circuit current. are the most remote (in electrical sense) points of the network, since it is these points that correspond to the smallest value of the single-phase short-circuit current.
The magnitude of the single-phase short-circuit current. can be determined by the approximate formula

where U f - phase voltage of the network, V;
Z T - total resistance of the step-down transformer to the fault current to the body, ohm;
Z P - loop impedance phase - line zero to the most remote point of the network, ohm.
The calculated values ​​of the impedances of step-down transformers for single-phase faults are given in table. 7-1.
For transformers with a power of more than 630 kva when determining the short-circuit current. can be accepted:
Z T =0
The total resistance of a loop of wires or cores of a line cable is determined by the formula

where R p - active phase resistance ( R f ) and neutral (Ro) wires, ohms;
R n =R f +R o (7-3)
X p - inductive reactance of a loop of wires or cable cores, ohms.

The active resistance of wires made of non-ferrous metals is determined according to table. 5-1. The average values ​​of inductive reactance of wire loops or cable cores made of non-ferrous metals per 1 km of line are given in Table. 7-2.
For steel wires, the inductive reactance of a wire loop is determined by the formula

where X"n - external inductive reactance of the loop of forward and return wires, equal to overhead line voltage up to 1000V 0.6 ohm/km; X" p.p and X" p.o - internal inductive resistances of the forward and return wires of the line, respectively, ohm/km.
The impedance values ​​of loops for wires and cable cores made of non-ferrous metals per 1 km of line are given in Table. 7-3. In table 7-6 shows the resistance of the loop “phase of a three-core cable - steel strip” for unarmored cables.

Table 7-1 Calculated resistance of transformers for single-phase short circuit. on the 400/230 V side

Type	Rated power, kVA	Voltage HV windings. kv	Connection diagram	Impedance Zt, ohm
GOST401-41
TM, TMA TM TM TM TMA TSMA TSM TM, TMA TM, TMA TM. TMA TM, TMA TM, TMA TM, TMA TM, TMA TM, TMA TM	20 30 50 100 100 100 100 180 180 320 320 560 560 750 1000 1000	6-10 6-10 6-10 6-10 35 6-10 35 6-10 35 6-10 35 6-10 35 6-10 6-10 35	U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un	1,39 0,9 0,54 0,27 0,25 0,26 0,25 0,15 0,14 0,085 0,08 0,048 0,046 0,036 0,027 0,026
GOST12022-66
TM TM TM TM TM TM TM TM TM TM TM TM TM TM	25 40 63 63 100 100 160 160 250 250 400 400 400 630	6-10 6-10 6-10 20 6-10 20-35 6-10 20-35 6-10 20-35 6-10 20-35 6-10 6-10	U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un U/Un D/Un U/Un	1,04 0,65 0,413 0,38 0,26 0,253 0,162 0,159 0,104 0,102 0,065 0,064 0,022 0,043
GOST11920-66
TM TM TM TM	1 000 1 000 1 000 1 000	6-10 20-35 6-10 20-35	U/Un U/Un D/Un D/Un	0,027 0,026 0,009 0,01
TSZ TSZ TSZ TSZ TSZ TSZ TSZ TSZ TSZ TSZ	160 180 250 320 400 560 630 750 1 000 1 000	6-10 6-10 6-10 6-10 6-10 6-10 6-10 6-10 6-10 6-10	D/Un U/Un D/Un U/Un D/Un U/Un D/Un U/Un D/Un U/Un	0,055 0,15 0,035 0,085 0,022 0,048 0,014 0,036 0,009 0,027

Notes: For step-down transformers with a secondary winding voltage of 230/133V, the resistance values ​​are 3 times less than those indicated in the table. 7-1.
Symbols for transformer connection diagrams:
U - star; Un - a star with a derived zero point; D - triangle.

Table 7-2 Average values ​​of inductive reactance of a loop of forward and return wires or cable cores made of non-ferrous metals ohm/km

Table 7-3 Loop impedances of the forward and return wires of the line or cable cores, ohm/km

Wire cross-section, mm.kv		Cable and wires in pipes		Wires on rollers and insulators		Overhead wires
direct	reverse	copper	aluminum	copper	aluminum	copper	aluminum
1 1,5 1,5 2,5 2,5 4 4 4 6 6 6 10 10 10 16 16 16 25 25 25 35 35 35 50 50 50 70 70 70 95 95 95 120 120 120 150 150 150	1 1 1,5 1,5 2,5 1,5 2,5 4 2,5 4 6 4 6 10 6 10 16 10 16 25 10 16 35 16 25 50 25 35 70 35 50 95 50 70 120 50 70 150	37,8 31,5 25,2 20,2 15,1 17,3 12,2 9,3 10,6 7,71 6,12 6,50 4,90 3,68 4,26 3,04 2,40 2,58 1,94 1,49 2,38 1,74 1,09 1,60 1,14 0,793 1,03 0,833 0,58 0,755 0,608 0,428 0,568 0,461 0,350 0,535 0,430 0,285	- - - - 25,2 - 20,5 15,8 17,9 13,2 10,5 11,1 8,42 6,32 7,24 5,14 3,96 4,44 3,26 2,56 4,08 2,90 1,84 2,62 1,92 1,29 1,74 1,39 0,932 1,27 0,99 0,797 0,922 0,745 0,561 0,862 0,687 0,446	- - 25,2 20,2 15,1 17,3 12,2 9,3 10,6 7,71 6,14 6,52 4,92 3,71 4,28 3,08 2,45 2,62 1,98 1,55 2,42 1,79 1,16 1,65 1,21 0,890 1,11 0,927 0,706 856 0,712 0,566 - - - - - -	- - - - 25,2 - 20,5 15,8 17,9 13,2 10,5 11,1 8,42 6,32 7,24 5,15 3,99 4,46 3,30 2,60 4,11 2,96 1,90 2,66 1,97 1,36 1,80 1,45 1,03 1,34 1,08 0,815 - - - - - -	- - - - - - - 9,3 - - 6,16 - 4,96 3,75 4,32 3,13 2,52 2,69 2,08 1,68 2,48 1,87 1,29 1,74 1,32 1,05 1,24 1,08 0,896 1,02 0,915 0,772 0,858 0,792 0,732 - - -	- - - - - - - - - - - - - - - - 4,03 4,50 3,34 2,66 4,15 3,00 1,96 2,70 2,03 1,44 1,86 1,53 1,13 1,42 1,18 0,907 1,09 0,945 0,808 1,04 0,808 0,732

Table 7-6 Loop impedances "phase of three-core cable - steel strip", ohm/km

Cable cross-section, mm.kv	Current and material of cable cores	Dimensions of steel strip, mm
		20X4		40X4		50X4	50X4	60X4		80X4		100X4, 100X6	100X5, 100X8
	The operating current of the maximum release of the machine, and		1400		1400		1400		1400		1400		1400
	Rated current fuse link of the instantaneous fuse, and
	Cable core material:	Loop impedance, ohm/ km
	Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum Copper Aluminum	9,59 13,52 7,76 10,34 6,36 7,86 5,6 6,49 5,14 5,70 4,91 5,30 4,75 5,02 4,64 4,83 4,57 4,70 4,51 4,62 4,47 4,56 4,44 4,52	8,42 12,35 6,59 9,17 5,19 6,69 4,43 5,32 3,97 4,53 3,74 4,13 3,58 3,85 3,47 3,66 3,40 3,53 3,34 3,45 3,30 3,39 3,27 3,35	7,82 11,79 5,97 8,59 4,55 6,07 3,78 4,68 3,31 3,88 3,09 3,48 2,92 3,19 2,81 3,0 2,73 2,87 2,69 2,8 2,65 2,74 2,63 2,7	7,45 11,42 6,60 8,22 4,18 5,7 3,41 4,31 2,94 3,51 2,71 3,11 2,55 2,72 2,44 2,63 2,36 2,50 2,32 2,43 2,28 2,37 2,26 2,33	7,40 11,37 5,54 8,17 4,11 5,63 3,32 4,24 2,86 3,43 2,64 3,03 2,47 2,74 2,37 2,55 2,29 2,42 2,24 2,35 2,21 2,29 2,18 2,25	7,17 11,14 5,31 7,94 3,98 5,4 3,09 3,01 2,63 3,2 2,4 2,8 2,24 2,5 2,4 2,32 2,06 2,19 2,01 2,12 1,98 2,06 1,95 2,02	7,14 11,13 5,27 7,92 3,83 5,37 3,04 3,96 2,57 3,15 2,35 2,74 2,19 2,45 2,08 2,26 2,01 2,14 1,96 2,07 1,93 2,01 1,90 1,96	6,92 10,91 5,05 7,7 3,61 5,15 2,82 3,74 2,35 2,93 2,13 2,52 1,97 2,23 1,86 2,04 1,79 1,92 1,74 1,85 1,71 1,79 1,68 1,74	6,82 10,81 4,95 7,61 3,5 5,05 2,71 3,64 2,24 2,82 2,01 2,41 1,86 2,12 1,75 1,93 1,67 1,8 1,63 1,74 1,60 1,65 1,58 1,64	6,59 10,58 4,72 7,38 3,27 4,82 2,48 3,41 2,01 2,59 1,78 2,18 1,63 1,89 1,52 1,7 1,44 1,57 1,4 1,51 1,37 1,47 1,35 1,41	6,56 10,56 4.68 7,34 3,22 4,77 2,42 3,36 1,95 2,53 1,73 2,12 1,57 1,83 1,46 1,64 1,38 1,51 1,35 1,45 1,31 1,39 1,28 1,35	6,45 10,45 4,57 7,23 3,1 4,66 2,31 3,25 1,84 2,42 1,62 2,01 1,46 1,72 1,35 1,53 1,27 1,40 1,24 1,34 1,2 1,28 1,17 1,24 I3 ; ITo- the smallest value of single-phase short-circuit current, determined by formula (7-1), a. Permissible multiple of the minimum short-circuit current. must be at least 3 in relation to the rated current of the fuse link and the rated current of the circuit breaker release having an inverse current characteristic, and not less than 1.1 KRin relation to the operating current of a circuit breaker having only an electromagnetic release (KR- coefficient taking into account the spread of characteristics of the release according to the factory data). For networks laid in explosive areas, the permissible short-circuit current ratios are: increase to a value of 4 in relation to the rated current of the fuse link and 6 in relation to the rated current of the circuit breaker with an inverse current characteristic. For networks protected only from short-circuit currents, in necessary cases (for example, for detuning from motor self-starting currents), it is allowed to overestimate the currents of fuse links and the settings of circuit breaker releases, but at the same time the short-circuit current multiplier is must have a value of at least 5 in relation to the rated current of the fuse link and at least 1.5 in relation to the operating current of the electromagnetic release of the machine. Values ​​of permissible short-circuit current multiplicity. for various network installation conditions are given in table. 7-8. Table 7-8 Values ​​of the permissible minimum short-circuit current ratio. in relation to the current of the protective device Laying conditions Permissible short-circuit current ratio in relation to to the rated current of the fuse link to the current setting of the circuit breaker,having only electromagnetic release (cut-off) to the rated current of the release circuit breaker with inverse current characteristic The network is laid in a non-explosive room, subject to the requirements of Table. 4-50 The network is laid in a non-explosive area, provided that the requirements of Table. 4-50 are not fulfilled The network is laid in an explosive area 3 1.1Kr 1,5 1.1Kr 3 Notes: KR- coefficient taking into account the spread of characteristics of automatic circuit breakers with electromagnetic releases. In the absence of factory data on the guaranteed accuracy of the operating current setting of a circuit breaker with an electromagnetic release (cut-off), it is allowed to take the value of the coefficient KRfor circuit breakers with rated current up to 100 A - equal to 1.4, above 100 A - equal to 1.25. If it is difficult to fulfill the requirements specified in table. 7-8, the use of fast-acting ground fault protection is allowed. Example 7-1. In Fig. Figure 7-1 shows a diagram of a four-wire overhead line made aluminum wires and receiving power from the distribution board buses 380/220 V. The system neutral is solidly grounded. The cross-sections of the wires and the lengths of the line sections are shown in Fig. 7-1. Neglecting the resistance of the external network to the switchboard busbars and the resistance of the transformer, check the operation of protective devices with a single-phase short circuit. at the most distant points of the line for the following options: 1. The line is protected by fuses with fuse links with a rated current of 80 A. 2. The line is protected automatic switch type A 3124 with combined releases for a rated current of 100 a. 3. The line is protected by a type A 3124 circuit breaker with electromagnetic releases with an operating current setting of 600 A. Rice. 7-1. Scheme for example WITHThe reasons for choosing one or another protection device are not considered here. The example has a limited purpose - to show typical cases checking protective shutdown for single-phase short circuit. Solution. We check the triggering condition of the protection devices using formula (7-5). We determine the resistance of the loop of the phase and neutral wires of the line with a single-phase short circuit. at the point for which the resistance value is greatest. According to the table 7-.3 we find the values ​​of the specific resistances of the “phase - zero” loop for the sections of the line sections: 3 X 70+1 X 35 Zn=1 .53 ohm/km; 3 X 35+1 X 16 Zn=3 , 0 ohm/ km; 3 X16 +1 X16 Zn=4 , 0 3 ohm/ km; We determine which of the points D or E is the calculated one. Loop resistance between points G and D 4.03 X 0.08=0.323 ohm; loop resistance between points G and E 3 X 0.13=0.39 ohm. The calculated point is E. The total resistance of the phase-zero loop between points A and E is: Zn= 1.53(0.07+0.08) +0.39 = 0.62 ohm. Rated phase voltage Un = 220 V. We determine the value of single-phase current during a short circuit. at the most remote point E of the network (according to the example conditions, Zt = 0 should be taken): We check that condition (7-5) is met for all three line protection options. Option 1. Permissible minimum short-circuit current ratio. in relation to the rated current of the fuse link according to table. 7-8 is equal to: TO 31 = 3. Hence: 3x80=240 a<355 а. This ensures reliable operation of the fuses protecting the line. Option 2. Permissible short-circuit current ratio in relation to the thermal element of the combined release, which has a characteristic inversely dependent on the current, is equal to: TO 31 = 3. Hence the ratio (7-5) 3x100=300 a<355 а performed. Option 3. According to the plant, the guaranteed setting accuracy for type A 3124 circuit breakers is ±15%. Taking in accordance with the instructions in the table. 7-8 safety factor equal to 1.1, we get: TO 31 = 1.1x1.15=1.27; 1.27x600=760 a>355 a. Reliability of operation of the circuit breaker during short circuit. at point E is not provided. Example 7-2. In a system with a solidly grounded neutral at a voltage of 380/220 V, the line is protected by fuses with fuse links for a rated current of 100 A. Assuming Zt = 0, determine the longest line length that will ensure reliable blowing of fuses in a single-phase short circuit. at the end of the line for the following line options: 1. Overhead line with aluminum wires with a cross section of 3 X 50+1 X 25 mm2. 2. Three-core cable with aluminum cores with a cross-section of 3X50 mm.sq. in an aluminum sheath, used as a grounding wire. 3. Three-core unarmored cable with aluminum conductors with a cross-section of 3 X 50 mm2 with a grounding bus in the form of a steel strip with a cross-section of 50 X 4 mm. Solution. According to the table 7-8 we determine the minimum permissible short-circuit current ratio: TO 31 = 3. The smallest permissible value of single-phase short-circuit current. Ik = 3x100=300 a. Considering that, according to the conditions of the example, Zt = 0, we find by formula (7-1) the maximum permissible “phase-zero” resistance of the line: We determine the resistivity of 1 km of the phase-zero loop: for option 1, according to table. 7-3 Zn = 2.03 ohm/km; for option 2 Zn = 1.03 ohm/km; for option 3 according to table. 7-6 Zn = 2.74 ohm/km. The maximum allowable line lengths will be: option 1 option 2 option 3 The longest line length is ensured by using a cable using an aluminum sheath as a grounding (neutral) wire.

A circuit consisting of a transformer phase and a circuit of phase and neutral conductors. The resistance of the phase-zero loop determines the current of such a short circuit.

If the phase-to-neutral loop resistance is high, it may turn out that the short circuit current is not sufficient to trigger the short circuit protection quickly. And the protection either does not turn off the short circuit at all, or turns off after a long time. All this time, dangerous voltage is present on the body of the electrical device.

In electrical installations up to 1000 V with neutral grounding, the safety of servicing electrical equipment in the event of a breakdown on the housing is ensured by disconnecting the damaged area with minimal time. When a phase wire is short-circuited to a neutral wire or equipment housing connected to the neutral of a transformer (or generator), a circuit is formed consisting of a transformer phase and a circuit of phase and neutral conductors. This circuit is usually called a “phase-zero” loop.

Checking the reliability and speed of disconnecting the damaged section of the network is as follows:

The short circuit current to the housing Is is determined. This current is compared with the calculated tripping current of the protection of the network section under test. If the emergency current possible in a given section of the network exceeds the protection operation current by a sufficient factor, the reliability of the shutdown is considered ensured.

Fig.5

R t, X t - active and inductive resistance of the secondary winding of the power transformer

R to - contact resistance of the contact connection

R a - resistance of protection and switching devices

R tt, X tt - active and inductive resistance of the secondary winding of the current transformer

R pr, X tpr - active and inductive resistance of the wire (the length of the wire in both cases is taken to be 80 m.)

Z - electrical receiver-consumer.

Inductive and active resistance of the transformer winding (mOhm)

Contact resistances are determined by the following formula

Phase-zero loop impedance

The calculated short-circuit current obtained We compare it with the response current of protective equipment. If the condition is met, the protection device will operate and its selection is made correctly

Let's calculate the resistance of the phase-zero loop

As a transformer we take the following

U LV =0.4 kV

R k =7.6 kV

We first determine the inductive and active resistance of the transformer winding (mOhm) using formulas (6.1) and (6.2)