PUE tables copper conductors wires cables. Permissible long-term currents for wires, cords and cables with rubber or plastic insulation. Wiring quality and aging

When repairing and designing electrical equipment, it becomes necessary to choose the right wires. You can use a special calculator or reference book. But for this you need to know the load parameters and cable laying features.

Why do you need to calculate the cable cross-section?

The following requirements apply to electrical networks:

safety;
reliability;
efficiency.

If the selected cross-sectional area of the wire is small, then the current loads on the cables and wires will be large, which will lead to overheating. As a result, an emergency may occur that will damage all electrical equipment and become dangerous to the life and health of people.

If you install wires with a large cross-sectional area, then safe use is ensured. But from a financial point of view there will be cost overruns. The correct choice of wire cross-section is the key to long-term safe operation and rational use of financial resources.

The cable cross-section is calculated based on power and current. Let's look at examples. To determine what wire cross-section is needed for 5 kW, you will need to use the PUE tables (“Electrical Installation Rules”). This directory is a regulatory document. It states that the choice of cable cross-section is made according to 4 criteria:

Supply voltage (single-phase or three-phase).
Conductor material.
Load current, measured in amperes (A), or power - in kilowatts (kW).
Cable location.

The PUE does not have a value of 5 kW, so you will have to choose the next larger value - 5.5 kW. For installation in an apartment today it is necessary to use copper wire. In most cases, installation is by air, so a cross-section of 2.5 mm² is suitable from the reference tables. In this case, the maximum permissible current load will be 25 A.

The above reference book also regulates the current for which the input circuit breaker (VA) is designed. According to the “Electrical Installation Rules”, with a load of 5.5 kW, the VA current should be 25 A. The document states that the rated current of the wire that is suitable for a house or apartment should be an order of magnitude greater than that of the VA. In this case, after 25 A there is 35 A. The last value must be taken as the calculated one. A current of 35 A corresponds to a cross section of 4 mm² and a power of 7.7 kW. So, the choice of the cross-section of the copper wire according to power is completed: 4 mm².

To find out what wire cross-section is needed for 10 kW, we will again use the reference book. If we consider the case for open wiring, then we need to decide on the cable material and the supply voltage. For example, for an aluminum wire and a voltage of 220 V, the nearest higher power will be 13 kW, the corresponding cross-section is 10 mm²; for 380 V the power will be 12 kW and the cross-section will be 4 mm².

Choose by power

Before choosing a cable cross-section based on power, you need to calculate its total value and make a list of electrical appliances located in the territory to which the cable is laid. The power must be indicated on each device; the corresponding units of measurement will be written next to it: W or kW (1 kW = 1000 W). Then you will need to add up the power of all equipment and get the total.

If you select a cable to connect one device, then only information about its energy consumption is sufficient. You can select wire cross-sections based on power in the PUE tables.

Table 1. Selection of wire cross-section based on power for cables with copper conductors

	For cable with copper conductors
	Voltage 220 V		Voltage 380 V
	Current, A	power, kWt	Current, A	power, kWt
1,5	19	4,1	16	10,5
2,5	27	5,9	25	16,5
4	38	8,3	30	19,8
6	46	10,1	40	26,4
10	70	15,4	50	33
16	85	18,7	75	49,5
25	115	25,3	90	59,4
35	135	29,7	115	75.9
50	175	38.5	145	95,7
70	215	47,3	180	118,8
95	260	57,2	220	145,2
120	300	66	260	171,6

Table 2. Selection of wire cross-section based on power for cables with aluminum conductors

Conductor cross-section, mm²	For cable with aluminum conductors
	Voltage 220 V		Voltage 380 V
	Current, A	power, kWt	Current, A	power, kWt
2,5	20	4,4	19	12,5
4	28	6,1	23	15,1
6	36	7,9	30	19,8
10	50	11,0	39	25,7
16	60	13,2	55	36,3
25	85	18,7	70	46,2
35	100	22,0	85	56,1
50	135	29,7	110	72,6
70	165	36,3	140	92,4
95	200	44,0	170	112,2
120	230	50,6	200	132,2

In addition, you need to know the network voltage: three-phase corresponds to 380 V, and single-phase - 220 V.

The PUE provides information for both aluminum and copper wires. Both have their advantages and disadvantages. Advantages of copper wires:

high strength;
elasticity;
oxidation resistance;
electrical conductivity is greater than that of aluminum.

The disadvantage of copper conductors is their high cost. In Soviet houses, aluminum electrical wiring was used during construction. Therefore, if a partial replacement occurs, it is advisable to install aluminum wires. The only exceptions are those cases when new wiring is installed instead of all the old wiring (up to the switchboard). Then it makes sense to use copper. It is unacceptable for copper and aluminum to come into direct contact, as this leads to oxidation. Therefore, a third metal is used to connect them.

You can independently calculate the wire cross-section according to power for a three-phase circuit. To do this, you need to use the formula: I=P/(U*1.73), where P – power, W; U – voltage, V; I – current, A. Then the cable cross-section is selected from the reference table depending on the calculated current. If the required value is not there, then the closest one is selected, which exceeds the calculated one.

How to calculate by current

The amount of current passing through a conductor depends on the length, width, resistivity of the latter and on temperature. When heated, the electric current decreases. Reference information is given for room temperature (18°C). To select the cable cross-section for current, use the PUE tables.

Table3. Electric current for copper wires and cords with rubber and PVC insulation

Conductor cross-sectional area, mm²
	open	in one pipe
	open	two single-core	three single-core	four single-core	one two-wire	one three-wire
0,5	11	-	-	-	-	-
0,75	15	-	-	-	-	-
1	17	16	15	14	15	14
1,2	20	18	16	15	16	14,5
1,5	23	19	17	16	18	15
2	26	24	22	20	23	19
2,5	30	27	25	25	25	21
3	34	32	28	26	28	24
4	41	38	35	30	32	27
5	46	42	39	34	37	31
6	50	46	42	40	40	34
8	62	54	51	46	48	43
10	80	70	60	50	55	50
16	100	85	80	75	80	70
25	140	115	100	90	100	85
35	170	135	125	115	125	100
50	215	185	170	150	160	135
70	270	225	210	185	195	175
95	330	275	255	225	245	215
120	385	315	290	260	295	250
150	440	360	330	-	-	-
185	510	-	-	-	-	-
240	605	-	-	-	-	-
300	695	-	-	-	-	-
400	830	-	-	-	-	-

A table is used to calculate aluminum wires.

Table4. Electric current for aluminum wires and cords with rubber and PVC insulation

Conductor cross-sectional area, mm²	Current, A, for wires laid
	open	in one pipe
	open	two single-core	three single-core	four single-core	one two-wire	one three-wire
2	21	19	18	15	17	14
2,5	24	20	19	19	19	16
3	27	24	22	21	22	18
4	32	28	28	23	25	21
5	36	32	30	27	28	24
6	39	36	32	30	31	26
8	46	43	40	37	38	32
10	60	50	47	39	42	38
16	75	60	60	55	60	55
25	105	85	80	70	75	65
35	130	100	95	85	95	75
50	165	140	130	120	125	105
70	210	175	165	140	150	135
95	255	215	200	175	190	165
120	295	245	220	200	230	190
150	340	275	255	-	-	-
185	390	-	-	-	-	-
240	465	-	-	-	-	-
300	535	-	-	-	-	-
400	645	-	-	-	-	-

In addition to the electric current, you will need to select the conductor material and voltage.

For an approximate calculation of the cable cross-section for current, it must be divided by 10. If the resulting cross-section is not in the table, then it is necessary to take the nearest larger value. This rule is only suitable for cases where the maximum permissible current for copper wires does not exceed 40 A. For the range from 40 to 80 A, the current must be divided by 8. If aluminum cables are installed, then it must be divided by 6. This is because for To ensure equal loads, the thickness of the aluminum conductor is greater than that of copper.

Calculation of cable cross-section by power and length

The cable length affects the voltage loss. Thus, at the end of the conductor the voltage may decrease and become insufficient for the operation of the electrical appliance. For household electrical networks, these losses can be neglected. It will be enough to take a cable 10-15 cm longer. This reserve will be used for switching and connection. If the ends of the wire are connected to the shield, then the spare length should be even greater, since circuit breakers will be connected.

When laying cables over long distances, voltage drop must be taken into account. Each conductor is characterized by electrical resistance. This parameter is affected by:

Wire length, unit of measurement – m. As it increases, losses increase.
Cross-sectional area, measured in mm². As it increases, the voltage drop decreases.
Material resistivity (reference value). Shows the resistance of a wire measuring 1 square millimeter per 1 meter.

The voltage drop is numerically equal to the product of resistance and current. It is acceptable that the specified value does not exceed 5%. Otherwise, you need to take a cable with a larger cross-section. Algorithm for calculating wire cross-section based on maximum power and length:

Depending on the power P, voltage U and cosph coefficient, we find the current using the formula: I=P/(U*cosph). For electrical networks that are used in everyday life, cosф = 1. In industry, cosф is calculated as the ratio of active power to total power. The latter consists of active and reactive powers.
Using PUE tables, the current cross-section of the wire is determined.
We calculate the resistance of the conductor using the formula: Ro=ρ*l/S, where ρ is the resistivity of the material, l is the length of the conductor, S is the cross-sectional area. It is necessary to take into account the fact that current flows through the cable not only in one direction, but also back. Therefore, the total resistance: R = Ro*2.
We find the voltage drop from the relationship: ΔU=I*R.
We determine the voltage drop as a percentage: ΔU/U. If the obtained value exceeds 5%, then select the nearest larger cross-section of the conductor from the reference book.

Open and closed wiring

Depending on the placement, wiring is divided into 2 types:

closed;
open.

Today, hidden wiring is installed in apartments. Special recesses are created in the walls and ceilings to accommodate cables. After installing the conductors, the recesses are plastered. Copper wires are used. Everything is planned in advance, because over time, to build up electrical wiring or replace elements, you will have to dismantle the finishing. For hidden finishing, wires and cables that have a flat shape are often used.

When laid open, the wires are installed along the surface of the room. Advantages are given to flexible conductors that have a round shape. They are easy to install in cable channels and pass through the corrugation. When calculating the load on the cable, the method of laying the wiring is taken into account.

Cable power table required to correctly calculate the cable cross-section, if the power of the equipment is large and the cable cross-section is small, then it will heat up, which will lead to the destruction of the insulation and loss of its properties.

To calculate the conductor resistance, you can use the conductor resistance calculator.

For the transmission and distribution of electric current, the main means are cables; they ensure the normal operation of everything related to electric current, and how good this work will be depends on the right choice cable cross-section by power. A convenient table will help you make the necessary selection:

Current cross-section conductive lived mm
	Voltage 220V		Voltage 380V
	Current. A	Power. kW	Current. A	Power, kWt

Section Toko- conductive lived mm	Aluminum conductors wires and cables
	Voltage 220V		Voltage 380V
	Current. A	Power. kW	Current. A	Power, kWt

But in order to use the table, you need to calculate the total power consumption of devices and equipment that are used in a house, apartment or other place where the cable will be laid.

Example of power calculation.

Let's say you are installing closed electrical wiring in a house using an explosive cable. You need to write down a list of equipment used on a piece of paper.

But how now find out power? You can find it on the equipment itself, where there is usually a label with the main characteristics recorded.

Power is measured in Watts (W, W) or Kilowatts (kW, KW). Now you need to write down the data and then add it up.

The resulting number is, for example, 20,000 W, which would be 20 kW. This figure shows how much energy all electrical receivers together consume. Next, you should consider how many devices will be used simultaneously over a long period of time. Let’s say it turns out to be 80%, in which case the simultaneity coefficient will be equal to 0.8. We calculate the cable cross-section based on power:

20 x 0.8 = 16 (kW)

To select a cross-section, you will need a cable power table:

Current cross-section conductive lived mm	Copper conductors of wires and cables
	Voltage 220V		Voltage 380V
	Current. A	Power. kW	Current. A	Power, kWt




10		15.4

If the three-phase circuit is 380 Volts, then the table will look like this:

Current cross-section conductive lived mm	Copper conductors of wires and cables
	Voltage 220V		Voltage 380V
	Current. A	Power. kW	Current. A	Power, kWt

				16.5


10		15.4

These calculations are not particularly difficult, but it is recommended to choose a wire or cable with the largest cross-section of conductors, because it may be that it will be necessary to connect some other device.

Additional cable power table.

When installing electrical wiring, it is necessary to determine the power of consumers in advance. This will help in the optimal selection of cables. This choice will allow you to operate the wiring for a long time and safely without repairs.

Cable and conductor products are very diverse in their properties and intended purpose, and also have a wide range in prices. The article talks about the most important wiring parameter - the cross-section of a wire or cable in terms of current and power, and how to determine the diameter - calculate it using a formula or select it using a table.

The current-carrying part of the cable is made of metal. The part of the plane passing at right angles to the wire, bounded by metal, is called wire cross-section. The unit of measurement is square millimeters.

Section determines permissible currents passing through wires and cables. This current, according to the Joule-Lenz law, leads to the release of heat (proportional to the resistance and the square of the current), which limits the current.

Conventionally, three temperature ranges can be distinguished:

the insulation remains intact;
the insulation burns, but the metal remains intact;
metal melts at high temperatures.

Of these, only the first is the permissible operating temperature. In addition, with a decrease in cross-section its electrical resistance increases, which leads to an increase in voltage drop in the wires.

However, an increase in cross-section leads to an increase in mass and especially cost or cable.

The materials used for the industrial production of cable products are pure copper or aluminum. These metals have different physical properties, in particular resistivity, and therefore the cross sections selected for a given current may be different.

Find out from this video how to choose the correct wire or cable cross-section according to power for home wiring:

Determination and calculation of cores using the formula

Now let’s figure out how to correctly calculate the cross-section of a wire based on power, knowing the formula. Here we will solve the problem of determining the section. It is the cross section that is the standard parameter due to the fact that the nomenclature includes both single-core and multi-core options. The advantage of multi-core cables is their greater flexibility and resistance to kinks during installation. As a rule, stranded wires are made of copper.

The easiest way to determine the cross-section of a round single-core wire is d– diameter, mm; S– area in square millimeters:

Stranded ones are calculated by a more general formula: n– number of veins, d– core diameter, S- square:

The diameter of the core can be determined by removing the insulation and measuring the diameter against the bare metal with a caliper or micrometer.

The current density is determined very simply, it is number of amperes per section. There are two wiring options: open and closed. The open one allows for a higher current density due to better heat transfer to the environment. Closed requires a downward adjustment so that the heat balance does not lead to overheating in the tray, cable duct or shaft, which can cause a short circuit or even a fire.

Accurate thermal calculations are very complex; in practice, they are based on the permissible operating temperature of the most critical element in the structure, according to which the current density is selected.

Thus, the permissible current density is the value at which heating of the insulation of all wires in a bundle (cable duct) remains safe, taking into account the maximum ambient temperature.

Table of current cross-section of copper and aluminum wire or cable:

Table 1 shows the permissible current density for temperatures not higher than room temperature. Most modern wires have PVC or polyethylene insulation, allowing heating during operation no more than 70-90°C. For “hot” rooms, the current density must be reduced by a factor of 0.9 for every 10°C to the operating temperatures of the wires or cables.

Now about what is considered open and what . is wiring if it is made with clamps (tires) along the walls, ceiling, along the supporting cable or through the air. The closed one is laid in cable trays, walled into walls under plaster, made in pipes, a shell or laid in the ground. You should also consider the wiring closed if it is in or. The closed one cools worse.

For example, let the thermometer in the dryer room show 50°C. To what value should the current density of a copper cable laid in this room along the ceiling be reduced if the cable insulation can withstand heating up to 90°C? The difference is 50-20 = 30 degrees, which means you need to use the coefficient three times. Answer:

Example of calculating the wiring section and load

Let the suspended ceiling be illuminated by six lamps with a power of 80 W each and they are already connected to each other. We need to supply power to them using aluminum cable. We will assume that the wiring is closed, the room is dry, and the temperature is room temperature. Now we’ll find out how to calculate the power of copper and aluminum cables; for this we use the equation that determines the power (according to the new standards, we consider the mains voltage to be equal to 230 V):

Using the corresponding current density for aluminum from Table 1, we find the cross section required for the line to operate without overheating:

If we need to find the diameter of the wire, we use the formula:

Suitable would be cable APPV2x1.5 (section 1.5 mm.kv). This is perhaps the thinnest cable you can find on the market (and one of the cheapest). In the above case, it provides a double power reserve, i.e. a consumer with a permissible load power of up to 500 W, for example, a fan, dryer or additional lamps, can be installed on this line.

It is unacceptable to install sockets on this line, since they may (and most likely will) contain a powerful consumer and this will lead to overloading the line section.

Quick selection: useful standards and ratios

To save time, calculations are usually tabulated, especially since the range of cable products is quite limited. The following table shows the calculation of the cross-section of copper and aluminum wires according to power consumption and current strength, depending on the purpose - for open and closed wiring. The diameter is obtained as a function of load power, metal and type of wiring. The mains voltage is considered to be 230 V.

The table allows you to quickly select a section or diameter, if the load power is known. The found value is rounded up to the nearest value from the nomenclature series.

The following table summarizes the data on permissible currents by cross-section and power of materials of cables and wires for calculation and quick selection of the most suitable ones:

The wiring arrangement, among other things, requires design skills, which not everyone who wants to do it has. It is not enough to just have good electrical installation skills. Some people confuse design with drawing up documentation according to some rules. These are completely different things. A good project can be written out on pieces of paper from a notebook.

First of all, draw a plan of your premises and mark future sockets and lamps. Find out the power of all your consumers: irons, lamps, heating devices, etc. Then enter the power of the loads most frequently consumed in different rooms. This will allow you to choose the best cable options.

You will be surprised how many possibilities there are and what is the reserve for saving money. Once you select , count the length of each line you draw. Put everything together, and then you will get exactly what you need, and as much as you need.

Each line must be protected by its own (), designed for a current corresponding to the permissible power of the line (the sum of the consumer powers). Sign the machines, located in, for example: “kitchen”, “living room”, etc.

It is advisable to have a separate line for all lighting, then you can easily repair the socket in the evening without using matches. It is the sockets that are most often overloaded. Provide outlets with enough power - you don't know in advance what you'll have to plug into them.

In damp rooms, use only double-insulated cables! Use modern sockets (“Euro”) and with grounding conductors and connect the grounding correctly. Bend single-core wires, especially copper ones, smoothly, leaving a radius of several centimeters. This will prevent them from breaking. Wires must lie straight in cable trays and ducts, but freely, in no case should you pull them like a string.

There should be a margin of a few extra centimeters. When laying, you need to make sure that there are no sharp corners anywhere that could cut the insulation. The terminals must be tightened tightly when connecting., and for stranded wires this procedure should be repeated; they have a tendency for the cores to shrink, as a result of which the connection may become loose.

Copper wires and aluminum wires are not “friendly” with each other for electrochemical reasons; they cannot be connected directly. To do this, you can use special terminal blocks or galvanized washers. The joints must always be dry.

Phase conductors must be white (or brown), and neutrals must always be blue. Grounding is yellow-green in color. These are generally accepted color rules and commercial cables, as a rule, have internal insulation in exactly these colors. Compliance with colors increases the safety of operation and repair.

We bring to your attention an interesting and educational video on how to correctly calculate the cable cross-section by power and length:

The choice of wire cross-section is the main element of a power supply project of any scale, from a room to large networks. The current that can be taken into the load and power will depend on this. The correct choice of wires also ensures electrical and fire safety, and provides an economical budget for your project.

1.3.1. This chapter of the Rules applies to the selection of cross-sections of electrical conductors (bare and insulated wires, cables and buses) for heating, economic current density and corona conditions. If the cross-section of the conductor determined according to these conditions is less than the cross-section required by other conditions (thermal and electrodynamic resistance to short-circuit currents, voltage losses and deviations, mechanical strength, overload protection), then the largest cross-section required by these conditions should be accepted.

Selection of heating conductor cross-sections

1.3.2. Conductors for any purpose must meet the requirements for maximum permissible heating, taking into account not only normal, but also post-emergency conditions, as well as conditions during repairs and possible uneven current distribution between lines, bus sections, etc. When checking for heating, a half-hour maximum is accepted current, the largest of the average half-hour currents of a given network element.

1.3.3. For intermittent and short-term operating modes of electrical receivers (with a total cycle duration of up to 10 minutes and an operating period of no more than 4 minutes), the current reduced to the long-term mode should be taken as the calculated current for checking the cross-section of heating conductors. Wherein:

1) for copper conductors with a cross-section of up to 6 mm², and for aluminum conductors up to 10 mm², the current is taken as for installations with long-term operation;

2) for copper conductors with a cross-section of more than 6 mm², and for aluminum conductors with a cross-section of more than 10 mm², the current is determined by multiplying the permissible continuous current by the coefficient, where Tpk- the duration of the working period expressed in relative units (the duration of switching on in relation to the duration of the cycle).

1.3.4. For a short-term operating mode with a switching duration of no more than 4 minutes and breaks between switching on sufficient to cool the conductors to ambient temperature, the maximum permissible currents should be determined according to the standards for repeated short-term duty (see 1.3.3). When the duration of switching on is more than 4 minutes, as well as during breaks of insufficient duration between switching on, the maximum permissible currents should be determined as for installations with a long operating mode.

1.3.5. For cables with voltages up to 10 kV with impregnated paper insulation that carry less than rated loads, a short-term overload indicated in table may be allowed. 1.3.1.

1.3.6. For the period of liquidation of the post-emergency regime, an overload of up to 10% is allowed for cables with polyethylene insulation, and for cables with polyvinyl chloride insulation up to 15% of the rated load during maximum loads lasting no more than 6 hours per day for 5 days, if the load during the remaining periods of time of these days does not exceed the nominal.

During the period of liquidation of the post-emergency regime, overloads are allowed for 5 days for cables with voltages up to 10 kV with paper insulation. within the limits specified in table. 1.3.2.

Table 1.3.1. Permissible short-term overload for cables with voltages up to 10 kV with impregnated paper insulation

Table 1.3.2. Permissible overload for the period of post-emergency liquidation for cables with voltage up to 10 kV with paper insulation

For cable lines that have been in operation for more than 15 years, overloads should be reduced by 10%.

Overloading cable lines with a voltage of 20-35 kV is not allowed.

1.3.7. The requirements for normal loads and post-accident overloads apply to cables and the connecting and termination couplings and terminations installed on them.

1.3.8. Zero working conductors in a four-wire three-phase current system must have a conductivity of at least 50% of the conductivity of the phase conductors; if necessary, it should be increased to 100% of the conductivity of the phase conductors.

1.3.9. When determining permissible long-term currents for cables, bare and insulated wires and busbars, as well as for rigid and flexible conductors laid in an environment whose temperature differs significantly from that given in 1.3.12-1.3.15 and 1.3.22, coefficients should be applied, given in table. 1.3.3.

Table 1.3.3. Correction factors for currents for cables, bare and insulated wires and busbars depending on ground and air temperatures

Conditional ambient temperature, °C	Standardized core temperature, °C	Correction factors for currents at the design temperature of the environment, °C
Conditional ambient temperature, °C	Standardized core temperature, °C	-5 and below	0	+5	+10	+15	+20	+25	+30	+35	+40	+45	+50
15	80	1,14	1,11	1,08	1,04	1,00	0,96	0,92	0,88	0,83	0,78	0,73	0,68
25	80	1,24	1,20	1,17	1,13	1,09	1,04	1,00	0,95	0,90	0,85	0,80	0,74
25	70	1,29	1,24	1,20	1,15	1,11	1,05	1,00	0,94	0,88	0,81	0,74	0,67
15	65	1,18	1,14	1,10	1,05	1,00	0,95	0,89	0,84	0,77	0,71	0,63	0,55
25	65	1,32	1,27	1,22	1,17	1,12	1,06	1,00	0,94	0,87	0,79	0,71	0,61
15	60	1,20	1,15	1,12	1,06	1,00	0,94	0,88	0,82	0,75	0,67	0,57	0,47
25	60	1,36	1,31	1,25	1,20	1,13	1,07	1,00	0,93	0,85	0,76	0,66	0,54
15	55	1,22	1,17	1,12	1,07	1,00	0,93	0,86	0,79	0,71	0,61	0,50	0,36
25	55	1,41	1,35	1,29	1,23	1,15	1,08	1,00	0,91	0,82	0,71	0,58	0,41
15	50	1,25	1,20	1,14	1,07	1,00	0,93	0,84	0,76	0,66	0,54	0,37	-
25	50	1,48	1,41	1,34	1,26	1,18	1,09	1,00	0,89	0,78	0,63	0,45	-

Permissible long-term stress for wires, cords and cables with rubber or plastic insulation

1.3.10. Permissible long-term currents for wires with rubber or polyvinyl chloride insulation, cords with rubber insulation and cables with rubber or plastic insulation in lead, polyvinyl chloride and rubber sheaths are given in Table. 1.3.4-1.3.11. They are accepted for temperatures: cores +65, ambient air +25 and ground + 15°C.

When determining the number of wires laid in one pipe (or cores of a stranded conductor), the neutral working conductor of a four-wire three-phase current system, as well as grounding and neutral protective conductors are not taken into account.

Permissible long-term currents for wires and cables laid in boxes, as well as in trays in bundles, must be accepted: for wires - according to table. 1.3.4 and 1.3.5 as for wires laid in pipes, for cables - according to table. 1.3.6-1.3.8 as for cables laid in the air. If the number of simultaneously loaded wires is more than four, laid in pipes, boxes, and also in trays in bundles, the currents for the wires should be taken according to the table. 1.3.4 and 1.3.5 as for wires laid openly (in the air), with the introduction of reduction factors of 0.68 for 5 and 6; 0.63 for 7-9 and 0.6 for 10-12 conductors.

For secondary circuit wires, reduction factors are not introduced.

Table 1.3.4. Permissible continuous current for wires and cords with rubber and polyvinyl chloride insulation with copper conductors


	open	in one pipe
	open	two single-core	three single-core	four single-core	one two-wire	one three-wire
0,5	11	-	-	-	-	-
0,75	15	-	-	-	-	-
1	17	16	15	14	15	14
1,2	20	18	16	15	16	14,5
1,5	23	19	17	16	18	15
2	26	24	22	20	23	19
2,5	30	27	25	25	25	21
3	34	32	28	26	28	24
4	41	38	35	30	32	27
5	46	42	39	34	37	31
6	50	46	42	40	40	34
8	62	54	51	46	48	43
10	80	70	60	50	55	50
16	100	85	80	75	80	70
25	140	115	100	90	100	85
35	170	135	125	115	125	100
50	215	185	170	150	160	135
70	270	225	210	185	195	175
95	330	275	255	225	245	215
120	385	315	290	260	295	250
150	440	360	330	-	-	-
185	510	-	-	-	-	-
240	605	-	-	-	-	-
300	695	-	-	-	-	-
400	830	-	-	-	-	-

Table 1.3.5. Permissible continuous current for rubber and polyvinyl chloride insulated wires with aluminum conductors

Conductor cross-section, mm²	Current, A, for wires laid
	open	in one pipe
	open	two single-core	three single-core	four single-core	one two-wire	one three-wire
2	21	19	18	15	17	14
2,5	24	20	19	19	19	16
3	27	24	22	21	22	18
4	32	28	28	23	25	21
5	36	32	30	27	28	24
6	39	36	32	30	31	26
8	46	43	40	37	38	32
10	60	50	47	39	42	38
16	75	60	60	55	60	55
25	105	85	80	70	75	65
35	130	100	95	85	95	75
50	165	140	130	120	125	105
70	210	175	165	140	150	135
95	255	215	200	175	190	165
120	295	245	220	200	230	190
150	340	275	255	-	-	-
185	390	-	-	-	-	-
240	465	-	-	-	-	-
300	535	-	-	-	-	-
400	645	-	-	-	-	-

Table 1.3.6. Permissible continuous current for wires with copper conductors with rubber insulation in metal protective sheaths and cables with copper conductors with rubber insulation in lead, polyvinyl chloride, nayrite or rubber sheaths, armored and unarmored

Conductor cross-section, mm²	Current *, A, for wires and cables
	single-core	two-wire		three-wire
	when laying
	in the air	in the air	in the ground	in the air	in the ground
__________________ * Currents apply to wires and cables both with and without a neutral core.
1,5	23	19	33	19	27
2,5	30	27	44	25	38
4	41	38	55	35	49
6	50	50	70	42	60
10	80	70	105	55	90
16	100	90	135	75	115
25	140	115	175	95	150
35	170	140	210	120	180
50	215	175	265	145	225
70	270	215	320	180	275
95	325	260	385	220	330
120	385	300	445	260	385
150	440	350	505	305	435
185	510	405	570	350	500
240	605	-	-	-	-

Table 1.3.7. Permissible continuous current for cables with aluminum conductors with rubber or plastic insulation in lead, polyvinyl chloride and rubber sheaths, armored and unarmored

Conductor cross-section, mm²	Current, A, for cables
	single-core	two-wire		three-wire
	when laying
	in the air	in the air	in the ground	in the air	in the ground
2,5	23	21	34	19	29
4	31	29	42	27	38
6	38	38	55	32	46
10	60	55	80	42	70
16	75	70	105	60	90
25	105	90	135	75	115
35	130	105	160	90	140
50	165	135	205	110	175
70	210	165	245	140	210
95	250	200	295	170	255
120	295	230	340	200	295
150	340	270	390	235	335
185	390	310	440	270	385
240	465	-	-	-	-

Note. Permissible continuous currents for four-core cables with plastic insulation for voltages up to 1 kV can be selected according to table. 1.3.7, as for three-core cables, but with a coefficient of 0.92.

Table 1.3.8. Permissible continuous current for portable light and medium hose cords, portable heavy duty hose cables, mine flexible hose cables, floodlight cables and portable wires with copper conductors

Conductor cross-section, mm²	Current *, A, for cords, wires and cables
Conductor cross-section, mm²	single-core	two-wire	three-wire
__________________ * Currents apply to cords, wires and cables with and without a neutral core.
0,5	-	12	-
0,75	-	16	14
1,0	-	18	16
1,5	-	23	20
2,5	40	33	28
4	50	43	36
6	. 65	55	45
10	90	75	60
16	120	95	80
25	160	125	105
35	190	150	130
50	235	185	160
70	290	235	200

Table 1.3.9. Permissible continuous current for portable hose cables with copper conductors and rubber insulation for peat enterprises

Table 1.3.10. Permissible continuous current for hose cables with copper conductors and rubber insulation for mobile electrical receivers

Table 1.3.11. Permissible continuous current for wires with copper conductors with rubber insulation for electrified transport 1.3 and 4 kV

Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A
1	20	16	115	120	390
1,5	25	25	150	150	445
2,5	40	35	185	185	505
4	50	50	230	240	590
6	65	70	285	300	670
10	90	95	340	350	745

Table 1.3.12. Reduction factor for wires and cables laid in boxes

Laying method	Number of laid wires and cables		Reduction factor for power supply wires
Laying method	single-core	stranded	separate electrical receivers with a utilization factor of up to 0.7	groups of electrical receivers and individual receivers with a utilization factor of more than 0.7
Multilayered and bundled	-	Up to 4	1,0	-
	2	5-6	0,85	-
	3-9	7-9	0,75	-
	10-11	10-11	0,7	-
	12-14	12-14	0,65	-
	15-18	15-18	0,6	-
Single layer	2-4	2-4	-	0,67
Single layer	5	5	-	0,6

1.3.11. Permissible long-term currents for wires laid in trays, when laid single-row (not in bundles), should be taken as for wires laid in the air.

Permissible long-term currents for wires and cables laid in boxes should be taken according to table. 1.3.4-1.3.7 as for single wires and cables laid openly (in the air), using the reduction factors indicated in table. 1.3.12.

When choosing reduction factors, control and reserve wires and cables are not taken into account.

Permissible continuous currents for cables with impregnated paper insulation

1.3.12. Permissible continuous currents for cables with voltages up to 35 kV with insulation made of impregnated cable paper in a lead, aluminum or polyvinyl chloride sheath are accepted in accordance with the permissible temperatures of the cable cores:

1.3.13. For cables laid in the ground, permissible long-term currents are given in table. 1.3.13, 1.3.16, 1.3.19-1.3.22. They are taken on the basis of laying no more than one cable in a trench at a depth of 0.7-1.0 m at a ground temperature of +15°C and a ground resistivity of 120 cm K/W.

Table 1.3.13. Permissible long-term current for cables with copper conductors with paper impregnated with oil rosin and non-drip insulation in a lead sheath, laid in the ground

Conductor cross-section, mm²	Current, A, for cables
	single-core up to 1 kV	two-wire up to 1 kV	three-wire voltage, kV			four-wire up to 1 kV
	single-core up to 1 kV	two-wire up to 1 kV	until 3	6	10	four-wire up to 1 kV
6	-	80	70	-	-	-
10	140	105	95	80	-	85
16	175	140	120	105	95	115
25	235	185	160	135	120	150
35	285	225	190	160	150	175
50	360	270	235	200	180	215
70	440	325	285	245	215	265
95	520	380	340	295	265	310
120	595	435	390	340	310	350
150	675	500	435	390	355	395
185	755	-	490	440	400	450
240	880	-	570	510	460	-
300	1000	-	-	-	-	-
400	1220	-	-	-	-	-
500	1400	-	-	-	-	-
625	1520	-	-	-	-	-
800	1700	-	-	-	-	-

Table 1.3.14. Permissible continuous current for cables with copper conductors with paper impregnated with oil rosin and non-drip insulation in a lead sheath, laid in water

Conductor cross-section, mm²	Current, A, for cables
	three-wire voltage, kV			four-wire up to 1 kV
	until 3	6	10	four-wire up to 1 kV
16	-	135	120	-
25	210	170	150	195
35	250	205	180	230
50	305	255	220	285
70	375	310	275	350
95	440	375	340	410
120	505	430	395	470
150	565	500	450	-
185	615	545	510	-
240	715	625	585	-

Table 1.3.15. Permissible continuous current for cables with copper conductors with paper impregnated with oil rosin and non-drip insulation in a lead sheath, laid in the air

Conductor cross-section, mm²	Current, A, for cables
	single-core up to 1 kV	two-wire up to 1 kV	three-wire voltage, kV			four-wire up to 1 kV
	single-core up to 1 kV	two-wire up to 1 kV	until 3	6	10	four-wire up to 1 kV
6	-	55	45	-	-	-
10	95	75	60	55	-	60
16	120	95	80	65	60	80
25	160	130	105	90	85	100
35	200	150	125	110	105	120
50	245	185	155	145	135	145
70	305	225	200	175	165	185
95	360	275	245	215	200	215
120	415	320	285	250	240	260
150	470	375	330	290	270	300
185	525	-	375	325	305	340
240	610	-	430	375	350	-
300	720	-	-	-	-	-
400	880	-	-	-	-	-
500	1020	-	-	-	-	-
625	1180	-	-	-	-	-
800	1400	-	-	-	-	-

Table 1.3.16. Permissible continuous current for cables with aluminum conductors with paper impregnated with oil rosin and non-drip insulation in a lead or aluminum sheath, laid in the ground

Conductor cross-section, mm²	Current, A, for cables
	single-core up to 1 kV	two-wire up to 1 kV	three-wire voltage, kV			four-wire up to 1 kV
	single-core up to 1 kV	two-wire up to 1 kV	until 3	6	10	four-wire up to 1 kV
6	-	60	55	-	-	-
10	110	80	75	60	-	65
16	135	110	90	80	75	90
25	180	140	125	105	90	115
35	220	175	145	125	115	135
50	275	210	180	155	140	165
70	340	250	220	190	165	200
95	400	290	260	225	205	240
120	460	335	300	260	240	270
150	520	385	335	300	275	305
185	580	-	380	340	310	345
240	675	-	440	390	355	-
300	770	-	-	-	-	-
400	940	-	-	-	-	-
500	1080	-	-	-	-	-
625	1170	-	-	-	-	-
800	1310	-	-	-	-	-

Table 1.3.17. Permissible continuous current for cables with aluminum conductors with paper impregnated with oil rosin and non-drip insulation in a lead sheath, laid in water

Conductor cross-section, mm²	Current, A, for cables
	three-wire voltage, kV			four-wire up to 1 kV
	until 3	6	10	four-wire up to 1 kV
16	-	105	90	-
25	160	130	115	150
35	190	160	140	175
50	235	195	170	220
70	290	240	210	270
95	340	290	260	315
120	390	330	305	360
150	435	385	345	-
185	475	420	390	-
240	550	480	450	-

Table 1.3.18. Permissible continuous current for cables with aluminum conductors with paper impregnated with oil rosin and non-drip insulation in a lead or aluminum sheath, laid in the air

Conductor cross-section, mm²	Current, A, for cables
	single-core up to 1 kV	two-wire up to 1 kV	three-wire voltage, kV			four-wire up to 1 kV
	single-core up to 1 kV	two-wire up to 1 kV	until 3	6	10	four-wire up to 1 kV
6	-	42	35	-	-	-
10	75	55	46	42	-	45
16	90	75	60	50	46	60
25	125	100	80	70	65	75
35	155	115	95	85	80	95
50	190	140	120	110	105	110
70	235	175	155	135	130	140
95	275	210	190	165	155	165
120	320	245	220	190	185	200
150	360	290	255	225	210	230
185	405	-	290	250	235	260
240	470	-	330	290	270	-
300	555	-	-	-	-	-
400	675	-	-	-	-	-
500	785	-	-	-	-	-
625	910	-	-	-	-	-
800	1080	-	-	-	-	-

Table 1.3.19. Permissible continuous current for three-core cables with a voltage of 6 kV with copper conductors with lean insulation in a common lead sheath, laid in the ground and air

Table 1.3.20. Permissible continuous current for three-core cables with a voltage of 6 kV with aluminum conductors with lean insulation in a common lead sheath, laid in the ground and air

Table 1.3.21. Permissible long-term current for cables with separately leaded copper conductors with paper impregnated with oil rosin and non-drip insulation, laid in the ground, water, air

Conductor cross-section, mm²
	20			35
	when laying
	in the ground	in water	in the air	in the ground	in water	in the air
25	110	120	85	-	-	-
35	135	145	100	-	-	-
50	165	180	120	-	-	-
70	200	225	150	-	-	-
95	240	275	180	-	-	-
120	275	315	205	270	290	205
150	315	350	230	310	-	230
185	355	390	265	-	-	-

Table 1.3.22. Permissible long-term current for cables with separately leaded aluminum conductors with paper impregnated with oil-rosin and non-drip insulation, laid in the ground, water, air

Conductor cross-section, mm²	Current, A, for three-core cables with voltage, kV
	20			35
	when laying
	in the ground	in water	in the air	in the ground	in water	in the air
25	85	90	65	-	-	-
35	105	110	75	-	-	-
50	125	140	90	-	-	-
70	155	175	115	-	-	-
95	185	210	140	-	-	-
120	210	245	160	210	225	160
150	240	270	175	240	-	175
185	275	300	205	-	-	-

Table 1.3.23. Correction factor for permissible continuous current for cables laid in the ground, depending on the resistivity of the earth

If the earth resistivity differs from 120 cm K/W, it is necessary to apply the correction factors indicated in the table to the current loads indicated in the previously mentioned tables. 1.3.23.

1.3.14. For cables laid in water, permissible continuous currents are given in table. 1.3.14, 1.3.17, 1.3.21, 1.3.22. They are taken based on water temperature +15°C.

1.3.15. For cables laid in the air, inside and outside buildings, with any number of cables and an air temperature of +25°C, the permissible continuous currents are given in table. 1.3.15, 1.3.18-1.3.22, 1.3.24, 1.3.25.

1.3.16. Permissible long-term currents for single cables laid in pipes in the ground must be taken as for the same cables laid in the air, at a temperature equal to the temperature of the ground.

Table 1.3.24. Permissible continuous current for single-core cables with a copper conductor with paper impregnated with oil rosin and non-drip insulation in a lead sheath, unarmoured, laid in the air

Conductor cross-section, mm²
Conductor cross-section, mm²	until 3	20	35
__________________
10	85/-	-	-
16	120/-	-	-
25	145/-	105/110	-
35	170/-	125/135	-
50	215/-	155/165	-
70	260/-	185/205	-
95	305/-	220/255	-
120	330/-	245/290	240/265
150	360/-	270/330	265/300
185	385/-	290/360	285/335
240	435/-	320/395	315/380
300	460/-	350/425	340/420
400	485/-	370/450	-
500	505/-	-	-
625	525/-	-	-
800	550/-	-	-

1.3.17. When laying mixed cables, permissible long-term currents must be taken for the section of the route with the worst cooling conditions, if its length is more than 10 m. It is recommended to use cable inserts with a larger cross-section in these cases.

1.3.18. When laying several cables in the ground (including laying in pipes), the permissible continuous currents must be reduced by introducing the coefficients given in table. 1.3.26. This does not include redundant cables.

Laying multiple cables in the ground with clear distances between them of less than 100 mm is not recommended.

1.3.19. For oil- and gas-filled single-core armored cables, as well as other cables of new designs, permissible continuous currents are established by the manufacturers.

1.3.20. Permissible long-term currents for cables laid in blocks should be determined using the empirical formula

I = abcI0,

Where I0- permissible continuous current for a three-core cable with a voltage of 10 kV with copper or aluminum conductors, determined according to table. 1.3.27; a- coefficient selected according to the table. 1.3.28 depending on the cross-section and location of the cable in the block; b- coefficient selected depending on the cable voltage:

c- coefficient selected depending on the average daily load of the entire block:

	1	0,85	0,7
Coefficient c	1	1,07	1,16

Table 1.3.25. Permissible continuous current for single-core cables with an aluminum core with paper impregnated with oil rosin and non-drip insulation in a lead or aluminum sheath, unarmoured, laid in the air

	Current *, A, for cables with voltage, kV
	until 3	20	35
__________________ * The numerator indicates currents for cables located in the same plane with a clear distance of 35-125 mm, the denominator indicates currents for cables located closely in a triangle.
10	65/-	-	-
16	90/-	-	-
25	110/-	80/85	-
35	130/-	95/105	-
50	165/-	120/130	-
70	200/-	140/160	-
95	235/-	170/195	-
120	255/-	190/225	185/205
150	275/-	210/255	205/230
185	295/-	225/275	220/255
240	335/-	245/305	245/290
300	355/-	270/330	260/330
400	375/-	285/350	-
500	390/-	-	-
625	405/-	-	-
800	425/-	-	-

Table 1.3.26. Correction factor for the number of working cables lying nearby in the ground (in pipes or without pipes)

Table 1.3.27. Permissible continuous current for cables, kV with copper or aluminum conductors with a cross-section of 95 mm², laid in blocks

Group	Block configuration	Channel no.	Current I, And for cables
Group	Block configuration	Channel no.	copper	aluminum
I		1	191	147
II		2	173	133
II		3	167	129
III		2	154	119
IV		2	147	113
IV		3	138	106
V		2	143	110
		3	135	104
		4	131	101
VI		2	140	103
		3	132	102
		4	118	91
VII		2	136	105
		3	132	102
		4	119	92
VIII		2	135	104
		3	124	96
		4	104	80
IX		2	135	104
		3	118	91
		4	100	77
X		2	133	102
		3	116	90
		4	81	62
XI		2	129	99
		3	114	88
		4	79	55

Table 1.3.28. Correction factor a per cable cross section

Conductor cross-section, mm2	Coefficient for the channel number in the block
Conductor cross-section, mm2	1	2	3	4
25	0,44	0,46	0,47	0,51
35	0,54	0,57	0,57	0,60
50	0,67	0,69	0,69	0,71
70	0,81	0,84	0,84	0,85
95	1,00	1,00	1,00	1,00
120	1,14	1,13	1,13	1,12
150	1,33	1,30	1,29	1,26
185	1,50	1,46	1,45	1,38
240	1,78	1,70	1,68	1,55

Backup cables may be laid in unnumbered channels of the unit if they work when the working cables are disconnected.

1.3.21. Permissible continuous currents for cables laid in two parallel blocks of the same configuration must be reduced by multiplying by coefficients selected depending on the distance between the blocks:

Permissible continuous currents for bare wires and buses

1.3.22. Permissible continuous currents for bare wires and painted tires are given in table. 1.3.29-1.3.35. They are taken based on the permissible heating temperature of +70°C at an air temperature of +25°C.

For hollow aluminum wires of grades PA500 and PA600, the permissible continuous current should be taken:

Wire brand	PA500	Pa6000
	1340	1680

1.3.23. When rectangular busbars are arranged flat, the currents given in table. 1.3.33, must be reduced by 5% for tires with a stripe width of up to 60 mm and by 8% for tires with a stripe width of more than 60 mm.

1.3.24. When choosing buses of large sections, it is necessary to choose the most economical design solutions in terms of throughput, ensuring the least additional losses from the surface effect and the proximity effect and the best cooling conditions (reducing the number of strips in the package, rational design of the package, the use of profile tires, etc.) .

Table 1.3.29. Permissible continuous current for bare wires according to GOST 839-80

Nominal cross-section, mm²	Section (aluminium/steel), mm2	Current, A, for wire brands
		AS, ASKS, ASK, ASKP		M	A and automatic transmission		M	A and automatic transmission
		outdoors	indoors	outdoors			indoors
10	10/1,8	84	53	95		-	60		-
16	16/2,7	111	79	133		105	102		75
25	25/4,2	142	109	183		136	137		106
35	35/6,2	175	135	223		170	173		130
50	50/8	210	165	275		215	219		165
70	70/11	265	210	337		265	268		210
95	95/16	330	260	422		320	341		255
120	120/19	390	313	485		375	395		300
120	120/27	375	-	485		375	395		300
150	150/19	450	365	570		440	465		355
	150/24	450	365
	150/34	450	-
185	185/24	520	430	650		500	540		410
	185/29	510	425
	185/43	515	-
240	240/32	605	505	760		590	685		490
	240/39	610	505
	240/56	610	-
300	300/39	710	600	880		680	740		570
	300/48	690	585
	300/66	680	-
330	330/27	730	-	-		-	-		-
400	400/22	830	713	1050		815	895		690
	400/51	825	705
	400/64	860	-
500	500/27	960	830	-		980	-		820
500	500/64	945	815	-		980	-		820
600	600/72	1050	920	-		1100	-		955
700	700/86	1180	1040	-		-	-		-

Table 1.3.30. Permissible continuous current for round and tubular busbars

Diam, mm	Round tires		Copper pipes		Aluminum pipes		Steel pipes
	Current *, A		Int. and external dia., mm	Current, A	Int. and external dia., mm	Current, A	Conditional passage, mm	Thickness walls, mm	External diameter, mm	Alternating current, A
	copper	aluminum	Int. and external dia., mm	Current, A	Int. and external dia., mm	Current, A	Conditional passage, mm	Thickness walls, mm	External diameter, mm	without incision	with extended cut
__________________ * The numerator shows loads with alternating current, the denominator shows loads with direct current.
6	155/155	120/120	12/15	340	13/16	295	8	2,8	13,5	75	-
7	195/195	150/150	14/18	460	17/20	345	10	2,8	17,0	90	-
8	235/235	180/180	16/20	505	18/22	425	15	3,2	21.3	118	-
10	320/320	245/245	18/22	555	27/30	500	20	3,2	26,8	145	-
12	415/415	320/320	20/24	600	26/30	575	25	4,0	33,5	180	-
14	505/505	390/390	22/26	650	25/30	640	32	4,0	42,3	220	-
15	565/565	435/435	25/30	830	36/40	765	40	4,0	48,0	255	-
16	610/615	475/475	29/34	925	35/40	850	50	4,5	60,0	320	-
18	720/725	560/560	35/40	1100	40/45	935	65	4,5	75,5	390	-
19	780/785	605/610	40/45	1200	45/50	1040	80	4,5	88,5	455	-
20	835/840	650/655	45/50	1330	50/55	1150	100	5,0	114	670	770
21	900/905	695/700	49/55	1580	54/60	1340	125	5,5	140	800	890
22	955/965	740/745	53/60	1860	64/70	1545	150	5,5	165	900	1000
25	1140/1165	885/900	62/70	2295	74/80	1770	-	-	-	-	-
27	1270/1290	980/1000	72/80	2610	72/80	2035	-	-	-	-	-
28	1325/1360	1025/1050	75/85	3070	75/85	2400	-	-	-	-	-
30	1450/1490	1120/1155	90/95	2460	90/95	1925	-	-	-	-	-
35	1770/1865	1370/1450	95/100	3060	90/100	2840	-	-	-	-	-
38	1960/2100	1510/1620	-	-	-	-	-	-	-	-	-
40	2080/2260	1610/1750	-	-	-	-	-	-	-	-	-
42	2200/2430	1700/1870	-	-	-	-	-	-	-	-	-
45	2380/2670	1850/2060	-	-	-	-	-	-	-	-	-

Table 1.3.31. Permissible continuous current for rectangular busbars

Size, mm	Copper bars				Aluminum tires				Steel tires
	Current *, A, with the number of stripes per pole or phase								Size, mm	Current *, A
	1	2	3	4	1	2	3	4	Size, mm	Current *, A
__________________ * The numerator shows the values of alternating current, the denominator shows the values of direct current.
15x3	210	-	-	-	165	-	-	-	16x2.5	55/70
20x3	275	-	-	-	215	-	-	-	20x2.5	60/90
25x3	340	-	-	-	265	-	-	-	25x2.5	75/110
30x4	475	-	-	-	365/370	-	-	-	20x3	65/100
40x4	625	-/1090	-	-	480	-/855	-	-	25x3	80/120
40x5	700/705	-/1250	-	-	540/545	-/965	-	-	30x3	95/140
50x5	860/870	-/1525	-/1895	-	665/670	-/1180	-/1470	-	40x3	125/190
50x6	955/960	-/1700	-/2145	-	740/745	-/1315	-/1655	-	50x3	155/230
60x6	1125/1145	1740/1990	2240/2495	-	870/880	1350/1555	1720/1940	-	60x3	185/280
80x6	1480/1510	2110/2630	2720/3220	-	1150/1170	1630/2055	2100/2460	-	70x3	215/320
100x6	1810/1875	2470/3245	3170/3940	-	1425/1455	1935/2515	2500/3040	-	75x3	230/345
60x8	1320/1345	2160/2485	2790/3020	-	1025/1040	1680/1840	2180/2330	-	80x3	245/365
80x8	1690/1755	2620/3095	3370/3850	-	1320/1355	2040/2400	2620/2975	-	90x3	275/410
100x8	2080/2180	3060/3810	3930/4690	-	1625/1690	2390/2945	3050/3620	-	100x3	305/460
120x8	2400/2600	3400/4400	4340/5600	-	1900/2040	2650/3350	3380/4250	-	20x4	70/115
60x10	1475/1525	2560/2725	3300/3530	-	1155/1180	2010/2110	2650/2720	-	22x4	75/125
80x10	1900/1990	3100/3510	3990/4450	-	1480/1540	2410/2735	3100/3440	-	25x4	85/140
100x10	2310/2470	3610/4325	4650/5385	5300/ 6060	1820/1910	2860/3350	3650/4160	4150/ 4400	30x4	100/165
120x10	2650/2950	4100/5000	5200/6250	5900/ 6800	2070/2300	3200/3900	4100/4860	4650/ 5200	40x4	130/220
-									50x4	165/270
									60x4	195/325
									70x4	225/375
									80x4	260/430
									90x4	290/480
									100x4	325/535

Table 1.3.32. Permissible continuous current for uninsulated bronze and steel-bronze wires

Table 1.3.33. Permissible continuous current for bare steel wires

Wire brand	Current, A	Wire brand	Current, A
PSO-3	23	PS-25	60
PSO-3.5	26	PS-35	75
PSO-4	30	PS-50	90
PSO-5	35	PS-70	125
-		PS-95	135

Table 1.3.34. Permissible continuous current for four-lane buses with stripes arranged on the sides of a square ("hollow package")

Dimensions, mm				Cross section of a four-lane tire, mm²	Current, A, per bus package
h	b	h1	H	Cross section of a four-lane tire, mm²	copper	aluminum
80	8	140	157	2560	5750	4550
80	10	144	160	3200	6400	5100
100	8	160	185	3200	7000	5550
100	10	164	188	4000	7700	6200
120	10	184	216	4800	9050	7300

Table 1.3.35. Permissible continuous current for box-section busbars

Dimensions, mm				Cross section of one tire, mm²	Current, A, for two buses
a	b	c	r	Cross section of one tire, mm²	copper	aluminum
75	35	4	6	520	2730	-
75	35	5,5	6	695	3250	2670
100	45	4,5	8	775	3620	2820
100	45	6	8	1010	4300	3500
125	55	6,5	10	1370	5500	4640
150	65	7	10	1785	7000	5650
175	80	8	12	2440	8550	6430
200	90	10	14	3435	9900	7550
200	90	12	16	4040	10500	8830
225	105	12,5	16	4880	12500	10300
250	115	12,5	16	5450	-	10800

Selection of wire cross-section based on economic current density

1.3.25. Conductor cross-sections must be checked for economic current density. Economically feasible section S, mm², is determined from the relation

S = I / Jek,

Where I- calculated current per hour of maximum power system, A; Jack- normalized value of economic current density, A/mm², for given operating conditions, selected according to table. 1.3.36.

The section obtained as a result of the specified calculation is rounded to the nearest standard section. The calculated current is taken for normal operation, i.e. the increase in current in post-emergency and repair modes of the network is not taken into account.

1.3.26. The selection of wire cross-sections for direct and alternating current power lines with voltages of 330 kV and above, as well as interconnection lines and powerful rigid and flexible conductors operating with a large number of hours of maximum use, is made on the basis of technical and economic calculations.

1.3.27. An increase in the number of lines or circuits beyond what is required under the conditions of reliability of power supply in order to satisfy the economic current density is carried out on the basis of a technical and economic calculation. In this case, in order to avoid increasing the number of lines or circuits, it is allowed to exceed the normalized values given in table two times. 1.3.36.

1,6

Feasibility calculations should take into account all investments in an additional line, including equipment and switchgear chambers at both ends of the lines. The feasibility of increasing the line voltage should also be checked.

These guidelines should also be followed when replacing existing wires with wires of larger cross-section or when laying additional lines to ensure economical current density as the load increases. In these cases, the full cost of all work on dismantling and installing line equipment, including the cost of apparatus and materials, must also be taken into account.

1.3.28. The following are not subject to verification by economic current density:

networks of industrial enterprises and structures with voltage up to 1 kV with the number of hours of use of the maximum load of enterprises up to 4000-5000;

branches to individual electrical receivers with voltages up to 1 kV, as well as lighting networks of industrial enterprises, residential and public buildings;

busbars of electrical installations and busbars within open and closed switchgears of all voltages;

conductors going to resistors, starting rheostats, etc.;

networks of temporary structures, as well as devices with a service life of 3-5 years.

1.3.29. When using the table. 1.3.36 the following must be followed (see also 1.3.27):

1. At maximum load at night, the economic current density increases by 40%.

2. For insulated conductors with a cross-section of 16 mm² or less, the economic current density increases by 40%.

3. For lines of the same section with n branch loads, the economic current density at the beginning of the line can be increased by kp times, and kp determined from the expression

Where I1, I2, ..., In- loads of individual sections of the line; l1, l2, ..., ln- lengths of individual sections of the line; L- total line length.

4. When choosing conductor cross-sections for power supply n similar, mutually redundant electrical receivers (for example, water supply pumps, converter units, etc.), of which m are in operation at the same time, the economic current density can be increased against the values given in table. 1.3.36, in kn times where kn equals:

1.3.30. The cross-section of 35 kV overhead line wires in rural areas feeding step-down substations 35/6 - 10 kV with transformers with voltage regulation under load should be selected according to the economic current density. It is recommended to take the design load when choosing wire sections for a 5-year perspective, counting from the year the overhead line was put into operation. For 35 kV overhead lines intended for redundancy in 35 kV networks in rural areas, the minimum long-term permissible current wire cross-sections should be used, based on the provision of power to electricity consumers in post-emergency and repair modes.

1.3.31. The choice of economic cross-sections of overhead wires and cores of cable lines with intermediate power take-offs should be made for each section, based on the corresponding calculated currents of the sections. In this case, for neighboring sections it is allowed to take the same wire cross-section corresponding to the economic cross-section for the longest section, if the difference between the values of the economic cross-section for these sections is within one step on the scale of standard sections. The cross-sections of wires on branches up to 1 km in length are taken to be the same as on the overhead line from which the branch is made. With a longer branch length, the economic cross-section is determined by the design load of this branch.

1.3.32. For power lines with voltage 6-20 kV given in table. 1.3.36 current density values may be used only when they do not cause voltage deviations at electricity receivers beyond permissible limits, taking into account the applied means of voltage regulation and reactive power compensation.

CHECKING CONDUCTORS FOR CORONA AND RADIO INTERFERENCE

1.3.33. At voltages of 35 kV and above, conductors must be checked for the conditions of corona formation, taking into account the average annual values of density and air temperature at the height of the electrical installation above sea level, the reduced radius of the conductor, as well as the coefficient of roughness of the conductors.

In this case, the highest field strength at the surface of any of the conductors, determined at the average operating voltage, should be no more than 0.9 of the initial electric field strength, corresponding to the appearance of a common corona.

The test should be carried out in accordance with current guidelines.

In addition, conductors must be tested according to the permissible level of radio interference from corona.

The table shows power, current and cross sections of cables and wires, For calculations and selection of cables and wires, cable materials and electrical equipment.

The calculation used data from the PUE tables and active power formulas for single-phase and three-phase symmetrical loads.

Below are tables for cables and wires with copper and aluminum wire cores.

Table for selecting cable cross-section for current and power with copper conductors

	Copper conductors of wires and cables
	Voltage, 220 V		Voltage, 380 V
	current, A	power, kWt	current, A	power, kWt
1,5	19	4,1	16	10,5
2,5	27	5,9	25	16,5
4	38	8,3	30	19,8
6	46	10,1	40	26,4
10	70	15,4	50	33,0
16	85	18,7	75	49,5
25	115	25,3	90	59,4
35	135	29,7	115	75,9
50	175	38,5	145	95,7
70	215	47,3	180	118,8
95	260	57,2	220	145,2
120	300	66,0	260	171,6

Table for selecting cable cross-section for current and power with aluminum conductors

Cross-section of current-carrying conductor, mm 2	Aluminum conductors of wires and cables
	Voltage, 220 V		Voltage, 380 V
	current, A	power, kWt	current, A	power, kWt
2,5	20	4,4	19	12,5
4	28	6,1	23	15,1
6	36	7,9	30	19,8
10	50	11,0	39	25,7
16	60	13,2	55	36,3
25	85	18,7	70	46,2
35	100	22,0	85	56,1
50	135	29,7	110	72,6
70	165	36,3	140	92,4
95	200	44,0	170	112,2
120	230	50,6	200	132,0

Example of cable cross-section calculation

Task: to power the heating element with a power of W=4.75 kW with copper wire in the cable channel.
Current calculation: I = W/U. We know the voltage: 220 volts. According to the formula, the flowing current I = 4750/220 = 21.6 amperes.

We focus on copper wire, so we take the value of the diameter of the copper core from the table. In the 220V - copper conductors column we find a current value exceeding 21.6 amperes, this is a line with a value of 27 amperes. From the same line we take the cross-section of the conductive core, equal to 2.5 squares.

Calculation of the required cable cross-section based on the type of cable or wire

№	Number of veins section mm. Cables (wires)	Outer diameter mm.							Pipe diameter mm.		Acceptable long current (A) for wires and cables when laying:		Permissible continuous current for rectangular copper bars sections (A) PUE
№	Number of veins section mm. Cables (wires)	VVG	VVGng	KVVG	KVVGE	NYM	PV1	PV3	PVC (HDPE)	Met.tr. Du	in the air	in the ground	Section, tires mm	Number of buses per phase
1	1x0.75							2,7	16	20	15	15	Section, tires mm	1	2	3
2	1x1							2,8	16	20	17	17	15x3	210
3	1x1.5	5,4	5,4				3	3,2	16	20	23	33	20x3	275
4	1x2.5	5,4	5,7				3,5	3,6	16	20	30	44	25x3	340
5	1x4	6	6				4	4	16	20	41	55	30x4	475
6	1x6	6,5	6,5				5	5,5	16	20	50	70	40x4	625
7	1x10	7,8	7,8				5,5	6,2	20	20	80	105	40x5	700
8	1x16	9,9	9,9				7	8,2	20	20	100	135	50x5	860
9	1x25	11,5	11,5				9	10,5	32	32	140	175	50x6	955
10	1x35	12,6	12,6				10	11	32	32	170	210	60x6	1125	1740	2240
11	1x50	14,4	14,4				12,5	13,2	32	32	215	265	80x6	1480	2110	2720
12	1x70	16,4	16,4				14	14,8	40	40	270	320	100x6	1810	2470	3170
13	1x95	18,8	18,7				16	17	40	40	325	385	60x8	1320	2160	2790
14	1x120	20,4	20,4						50	50	385	445	80x8	1690	2620	3370
15	1x150	21,1	21,1						50	50	440	505	100x8	2080	3060	3930
16	1x185	24,7	24,7						50	50	510	570	120x8	2400	3400	4340
17	1x240	27,4	27,4						63	65	605		60x10	1475	2560	3300
18	3x1.5	9,6	9,2			9			20	20	19	27	80x10	1900	3100	3990
19	3x2.5	10,5	10,2			10,2			20	20	25	38	100x10	2310	3610	4650
20	3x4	11,2	11,2			11,9			25	25	35	49	120x10	2650	4100	5200
21	3x6	11,8	11,8			13			25	25	42	60	rectangular copper bars (A) Schneider Electric IP30
22	3x10	14,6	14,6						25	25	55	90
23	3x16	16,5	16,5						32	32	75	115
24	3x25	20,5	20,5						32	32	95	150
25	3x35	22,4	22,4						40	40	120	180	Section, tires mm	Number of buses per phase
26	4x1			8	9,5				16	20	14	14	Section, tires mm	1	2	3
27	4x1.5	9,8	9,8	9,2	10,1				20	20	19	27	50x5	650	1150
28	4x2.5	11,5	11,5	11,1	11,1				20	20	25	38	63x5	750	1350	1750
29	4x50	30	31,3						63	65	145	225	80x5	1000	1650	2150
30	4x70	31,6	36,4						80	80	180	275	100x5	1200	1900	2550
31	4x95	35,2	41,5						80	80	220	330	125x5	1350	2150	3200
32	4x120	38,8	45,6						100	100	260	385	Permissible continuous current for rectangular copper bars (A) Schneider Electric IP31
33	4x150	42,2	51,1						100	100	305	435
34	4x185	46,4	54,7						100	100	350	500
35	5x1			9,5	10,3				16	20	14	14
36	5x1.5	10	10	10	10,9	10,3			20	20	19	27	Section, tires mm	Number of buses per phase
37	5x2.5	11	11	11,1	11,5	12			20	20	25	38	Section, tires mm	1	2	3
38	5x4	12,8	12,8			14,9			25	25	35	49	50x5	600	1000
39	5x6	14,2	14,2			16,3			32	32	42	60	63x5	700	1150	1600
40	5x10	17,5	17,5			19,6			40	40	55	90	80x5	900	1450	1900
41	5x16	22	22			24,4			50	50	75	115	100x5	1050	1600	2200
42	5x25	26,8	26,8			29,4			63	65	95	150	125x5	1200	1950	2800
43	5x35	28,5	29,8						63	65	120	180
44	5x50	32,6	35						80	80	145	225
45	5x95	42,8							100	100	220	330
46	5x120	47,7							100	100	260	385
47	5x150	55,8							100	100	305	435
48	5x185	61,9							100	100	350	500
49	7x1			10	11				16	20	14	14
50	7x1.5			11,3	11,8				20	20	19	27
51	7x2.5			11,9	12,4				20	20	25	38
52	10x1			12,9	13,6				25	25	14	14
53	10x1.5			14,1	14,5				32	32	19	27
54	10x2.5			15,6	17,1				32	32	25	38
55	14x1			14,1	14,6				32	32	14	14
56	14x1.5			15,2	15,7				32	32	19	27
57	14x2.5			16,9	18,7				40	40	25	38
58	19x1			15,2	16,9				40	40	14	14
59	19x1.5			16,9	18,5				40	40	19	27
60	19x2.5			19,2	20,5				50	50	25	38
61	27x1			18	19,9				50	50	14	14
62	27x1.5			19,3	21,5				50	50	19	27
63	27x2.5			21,7	24,3				50	50	25	38
64	37x1			19,7	21,9				50	50	14	14
65	37x1.5			21,5	24,1				50	50	19	27
66	37x2.5			24,7	28,5				63	65	25	38

Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A
1	20	16	115	120	390
1,5	25	25	150	150	445
2,5	40	35	185	185	505
4	50	50	230	240	590
6	65	70	285	300	670
10	90	95	340	350	745

Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A
1	20	16	115	120	390
1,5	25	25	150	150	445
2,5	40	35	185	185	505
4	50	50	230	240	590
6	65	70	285	300	670
10	90	95	340	350	745

Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A	Conductor cross-section, mm²	Current, A
1	20	16	115	120	390
1,5	25	25	150	150	445
2,5	40	35	185	185	505
4	50	50	230	240	590
6	65	70	285	300	670
10	90	95	340	350	745