Pue cable cross section table. Long-term permissible current loads of cables. Causes of cable heating

04.10.2023

The most important topic when designing power supply is the choice of cables based on the rated current. I have touched on this topic more than once and many people know my position, some agree, some don’t, however, today I want to dig a little deeper...
And it all started with this:

In general, I decided to check the words of Alexander Shalygin. By the way, I must say that I am very grateful to Alexander for his answers to controversial design questions, however, sometimes I do not agree with him.
I have an article:
In it I recently posted Shalygin’s answer on choosing cables.

In the question and answer, only PUE and GOST R 50571.5.52-2011 are mentioned, not a word is said about GOST 31996-2012.

GOST 31996-2012 is a document that cable products must comply with. There are other documents, but we will not touch on them, because... We will check it using a PVC insulated cable as an example.

The main idea is that different documents give different current values due to different temperatures of air, ground, and ground resistivity.

ROV	Pace. lived	Pace. air	Pace. land	*Earth resistivity, Km/W**
PUE	+65	+25	+15	1,2
GOST R 50571.5.52-2011	+70	+30	+20	2,5
GOST 31996-2012	+70	+25	+15	1,2

The first thing that catches your eye is that the PUE and GOST 31996-2012 adopt the same temperatures of air, ground and ground resistivity. Therefore, these documents must contain the same long-term permissible currents.

The question is about the cable APvBShvng 4×120. In this case, the current is determined according to table 1.3.7 PUE. The PUE does not have a table at all for cables with cross-linked polyethylene insulation.

APvBShvng 4×120 - power cable with cross-linked polyethylene insulation, with armor, reduced fire hazard.

To make our experiment cleaner, let’s replace the APvBShvng 4×120 cable with AVBbShv 4×120 and look at the currents in different documents when laid in the ground.

If our formulas are the same, then why are different currents presented in PUE and GOST 31996-2012? Why didn’t our currents coincide until the third digit?

271.4-226.92=44.48A - which is about 16%.

Since in the PUE and GOST R 50571.5.52-2011 the currents are given for different conditions, let's try to bring the currents to the same conditions.

1 Let’s calculate the permissible current of the AVBbShv-4×120 cable when laid in the ground at a ground temperature of +15 degrees and a resistivity of 1.2 K*m/W according to GOST R 50571.5.52-2011.

According to table B.52.16, using the interpolation method, we determine the correction factor for a resistivity of 1.2 K*m/W:

169*1.412=238.6A – current taking into account the earth resistivity of 1.2 K*m/W.

However, we must assume the temperature of the earth is +15 degrees. According to table B.52.15 – correction factor 1.05. The only caveat is that this coefficient is for laying cables in pipes in the ground. In my opinion, when laying directly in the ground, we should take the same coefficient.

238.6*1.05=250.5A – current taking into account the ground temperature of +15 degrees.

271.4-250.5 = 20.9A - which is about 8%.

2 Let’s calculate the permissible current of the AVBBShV-4×120 cable when laid in the ground at a ground temperature of +20 degrees and a resistivity of 2.5 K*m/W according to the PUE.
According to Table 1.3.23, we will determine the correction factor by interpolation:

271.4*0.81=219.8A – current taking into account the earth resistivity of 2.5 K*m/W.

According to table 1.3.3, the correction factor is 0.95 at a ground temperature of +20 degrees.

219.8A*0.95=208.8A – current taking into account the ground temperature of +20 degrees.

208.8-169 = 39.8A - which is about 19%.

What did I want to show with this?

If we bring all the documents to the same conditions, then the PUE and GOST R 50571.5.52-2011 present higher permissible currents for cables and differ from GOST 31996-2012, thereby it is possible manipulate various documents when justifying the cable cross-section.

In practice, attention is rarely paid to the temperature of the air, the ground, and the resistivity of the ground. Perhaps somewhere in the north or in the hot tropics this needs to be taken more seriously.

If you select cables in accordance with GOST R 50571.5.52-2011, then our networks are more secure. Often we do not know the values of the resistivity of the earth, so we can use Shalygin’s recommendations.

Ideally, you need to know the resistivity of the earth in order to choose the right cable when it comes to laying cables in the ground. At the same time, you must understand that it is not so easy to increase the cable cross-section. For the designer it’s just a number, but for the customer it’s money, which he’s in no hurry to part with.

Almost always I choose cables in accordance with GOST 31996-2012, especially since GOST R 50571.5.52-2011 is not valid in the Republic of Belarus

Regulatory documents for determining the permissible current of cables:

1 Rules for electrical installations.

2 GOST R 50571.5.52-2011 (Selection and installation of electrical equipment. Electrical wiring).

3 GOST 31996-2012 (Power cables with plastic insulation for rated voltages of 0.66, 1 and 3 kV).

P.S. I hope I didn't mess anything up

Letter dated July 21, 2014 No. 10-00-12/1188 (ROSTEKHNADZOR)

On amendments to the Electrical Installation Rules

The choice of which document to follow (GOST or PUE) depends on the specific situation.
At the same time, we inform you that the need to use the above documents in specific conditions is determined by the designer, who is responsible for the improper preparation of technical documentation, including shortcomings during construction, as well as during the operation of the facility (Article 761 of the Civil Code).

To make calculations easier, you can use Miroshko Leonid’s calculator

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

	open (in tray)	1 + 1 (two 1zh)	1 + 1 + 1 (three 1zh)	1 + 1 + 1 + 1 (four 1zh)	1*2 (one 2f)	1*3 (one 3zh)
0,5	11	—	—	—	—	—
0,75	15	—	—	—	—	—
1,00	17	16	15	14	15	14
1,5	23	19	17	16	18	15
2,5	30	27	25	25	25	21
4,0	41	38	35	30	32	27
6,0	50	46	42	40	40	34
10,0	80	70	60	50	55	50
16,0	100	85	80	75	80	70
25,0	140	115	100	90	100	85
35,0	170	135	125	115	125	100
50,0	215	185	170	150	160	135
70,0	270	225	210	185	195	175
95,0	330	275	255	225	245	215
120,0	385	315	290	260	295	250
150,0	440	360	330	—	—	—
185,0	510	—	—	—	—	—
240,0	605	—	—	—	—	—
300,0	695	—	—	—	—	—
400,0	830	—	—	—	—	—
Conductor cross-section, mm 2	open (in tray)	1 + 1 (two 1zh)	1 + 1 + 1 (three 1zh)	1 + 1 + 1 + 1 (four 1zh)	1 * 2 (one 2f)	1 * 3 (one 3zh)
Conductor cross-section, mm 2	Current loads A of wires laid in one pipe (box, bundle)

PUE, Table 1.3.5. Permissible continuous current for wires with rubber and polyvinyl chloride insulation with aluminum conductors
Conductor cross-section, mm 2	Current loads A of wires laid in one pipe (box, bundle)
Conductor cross-section, mm 2	open (in tray)	1 + 1 (two 1zh)	1 + 1 + 1 (three 1zh)	1 + 1 + 1 + 1 (four 1zh)	1*2 (one 2f)	1*3 (one 3zh)
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	—	—	—	—	—
Conductor cross-section, mm 2	open (in tray)	1 + 1 (two 1zh)	1 + 1 + 1 (three 1zh)	1 + 1 + 1 + 1 (four 1zh)	1 * 2 (one 2f)	1 * 3 (one 3zh)
Conductor cross-section, mm 2	Current loads A of wires laid in one pipe (box, bundle)

PUE, 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, mm 2
	single-core	two-wire		three-wire
	when laying
	in the air	in the air	in the ground	in the air	in the ground
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	—	—	—	—

PUE, 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, mm 2	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
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	—	—	—	—

PUE, 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, mm 2	Current *, A, for wires and cables
Conductor cross-section, mm 2	single-core	two-wire	three-wire
0.5	—	12	—
0.75	—	16	14
1	—	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

GOST 16442-80, Table 23. Permissible current loads of cables up to 3KV inclusive. with copper conductors with insulation made of polyethylene and polyvinyl chloride plastic, A*
Conductor cross-section, mm 2	Current *, A, for wires and cables
	single-core		two-wire		three-wire
	when laying
	in the air	in the ground	in the air	in the ground	in the air	in the ground
1,5	29	32	24	33	21	28
2,5	40	42	33	44	28	37
4	53	54	44	56	37	48
6	67	67	56	71	49	58
10	91	89	76	94	66	77
16	121	116	101	123	87	100
25	160	148	134	157	115	130
35	197	178	166	190	141	158
50	247	217	208	230	177	192
70	318	265	—	—	226	237
95	386	314	—	—	274	280
120	450	358	—	—	321	321
150	521	406	—	—	370	363
185	594	455	—	—	421	406
240	704	525	—	—	499	468

GOST 16442-80, Table 24. Permissible current loads of cables up to 3KV inclusive. with aluminum conductors with insulation made of polyethylene and polyvinyl chloride plastic, A*
Conductor cross-section, mm 2	Current *, A, for wires and cables
	single-core		two-wire		three-wire
	when laying
	in the air	in the ground	in the air	in the ground	in the air	in the ground
2.5	30	32	25	33	51	28
4	40	41	34	43	29	37
6	51	52	43	54	37	44
10	69	68	58	72	50	59
16	93	83	77	94	67	77
25	122	113	103	120	88	100
35	151	136	127	145	106	121
50	189	166	159	176	136	147
70	233	200	—	—	167	178
95	284	237	—	—	204	212
120	330	269	—	—	236	241
150	380	305	—	—	273	278
185	436	343	—	—	313	308
240	515	396	—	—	369	355

* Currents apply to wires and cables both with and without a neutral core.

The sections are taken based on heating the cores to 65°C at an ambient temperature of +25°C. When determining the number of wires laid in one pipe, the neutral working wire of a four-wire three-phase current system (or grounding wire) is not included in the calculation.

Current loads for wires laid in trays (not in bundles) are the same as for wires laid openly.

If the number of simultaneously loaded conductors laid in pipes, boxes, and also in trays in bundles is more than four, then the cross-section of the conductors must be selected as for conductors laid openly, but with the introduction of reduction factors for the current: 0.68 for 5 and 6 conductors , 0.63 - at 7-9, 0.6 - at 10-12.

Comfort and safety in the home depends on the correct choice of electrical wiring cross-section. When overloaded, the conductor overheats and the insulation may melt, causing a fire or short circuit. But it is not profitable to take a cross-section larger than necessary, since the price of the cable increases.

In general, it is calculated depending on the number of consumers, for which they first determine the total power used by the apartment, and then multiply the result by 0.75. The PUE uses a table of loads along the cable cross-section. From it you can easily determine the diameter of the cores, which depends on the material and the passing current. As a rule, copper conductors are used.

The cross-section of the cable core must correspond exactly to the calculated one - in the direction of increasing the standard size range. It is most dangerous when it is underestimated. Then the conductor constantly overheats, and the insulation quickly fails. And if you install the appropriate one, it will trigger frequently.

If the wire cross-section is increased, it will cost more. Although a certain reserve is necessary, since in the future, as a rule, it is necessary to connect new equipment. It is advisable to use a safety factor of about 1.5.

Calculation of total power

The total power consumed by the apartment falls on the main input, which enters the distribution board, and after it branches into the lines:

lighting;
groups of sockets;
individual powerful electrical appliances.

Therefore, the largest cross-section of the power cable is at the input. On outlet lines it decreases, depending on the load. First of all, the total power of all loads is determined. This is not difficult, since it is indicated on the housings of all household appliances and in their passports.

All powers add up. Calculations are made similarly for each circuit. Experts suggest multiplying the amount by 0.75. This is explained by the fact that all devices are not connected to the network at the same time. Others suggest choosing a larger section. Due to this, a reserve is created for the subsequent commissioning of additional electrical devices that may be purchased in the future. It should be noted that this cable calculation option is more reliable.

How to determine the wire cross-section?

All calculations include the cable cross-section. It is easier to determine it by diameter if you use the formulas:

S=π D²/4;
D= √(4×S/π).

Where π = 3.14.

S = N×D²/1.27.

Stranded wires are used where flexibility is required. Cheaper solid conductors are used for permanent installations.

How to choose a cable based on power?

In order to select wiring, use the load table for the cable cross-section:

If the open type line is energized at 220 V, and the total power is 4 kW, a copper conductor with a cross section of 1.5 mm² is taken. This size is usually used for lighting wiring.
With a power of 6 kW, conductors of a larger cross-section are required - 2.5 mm². The wire is used for sockets to which household appliances are connected.
A power of 10 kW requires the use of 6 mm² wiring. It is usually intended for the kitchen, where an electric stove is connected. The supply to such a load is made through a separate line.

Which cables are better?

Electricians are well aware of the cable of the German brand NUM for office and residential premises. In Russia they produce brands of cables that have lower characteristics, although they may have the same name. They can be distinguished by compound leaks in the space between the cores or by its absence.

The wire is produced monolithic and multi-wire. Each core, as well as all twisting, is insulated on the outside with PVC, and the filler between them is non-flammable:

Thus, the NUM cable is used indoors, since the insulation outdoors is destroyed by sunlight.
And as an internal cable, VVG brand cable is widely used. It is cheap and quite reliable. It is not recommended to use it for laying in the ground.
VVG brand wire is made flat and round. No filler is used between the cores.
made with an outer shell that does not support combustion. The cores are manufactured round up to a cross-section of 16 mm², and above - sector.
The PVS and ShVVP cable brands are made multi-wire and are used primarily for connecting household appliances. It is often used as home electrical wiring. It is not recommended to use multi-wire conductors outdoors due to corrosion. In addition, bending insulation will crack at low temperatures.
On the street, armored and moisture-resistant cables AVBShv and VBShv are laid underground. The armor is made of two steel strips, which increases the reliability of the cable and makes it resistant to mechanical stress.

Determination of current load

A more accurate result is obtained by calculating the cable cross-section by power and current, where the geometric parameters are related to the electrical ones.

For home wiring, not only the active load must be taken into account, but also the reactive load. The current strength is determined by the formula:

I = P/(U∙cosφ).

The reactive load is created by fluorescent lamps and motors of electrical appliances (refrigerator, vacuum cleaner, power tools, etc.).

Current example

Let's find out what to do if it is necessary to determine the cross-section of a copper cable for connecting household appliances with a total power of 25 kW and three-phase machines with a capacity of 10 kW. This connection is made with a five-core cable laid in the ground. The food at home comes from

Taking into account the reactive component, the power of household appliances and equipment will be:

P everyday life = 25/0.7 = 35.7 kW;
P rev. = 10/0.7 = 14.3 kW.

The input currents are determined:

I life = 35.7 × 1000/220 = 162 A;
I rev. = 14.3×1000/380 = 38 A.

If single-phase loads are distributed evenly across three phases, one will carry the current:

I f = 162/3 = 54 A.

I f = 54 + 38 = 92 A.

All equipment will not work at the same time. Taking into account the reserve, each phase accounts for the current:

I f = 92×0.75×1.5 = 103.5 A.

In a five-core cable, only the phase conductors are taken into account. For a cable laid in the ground, you can determine a core cross-section of 16 mm² for a current of 103.5 A (table of loads by cable cross-section).

Refined calculation of current allows you to save money, since a smaller cross-section is required. With a rougher calculation of cable power, the core cross-section will be 25 mm², which will cost more.

Cable voltage drop

Conductors have resistance that must be taken into account. This is especially important for long cable lengths or small cross-sections. PES standards have been established, according to which the voltage drop on the cable should not exceed 5%. The calculation is done as follows.

The conductor resistance is determined: R = 2×(ρ×L)/S.
Voltage drop is found: U pad. = I×R. In relation to the linear percentage, it will be: U % = (U falling / U linear) × 100.

The following notations are used in the formulas:

ρ - resistivity, Ohm×mm²/m;
S - cross-sectional area, mm².

Coefficient 2 shows that current flows through two wires.

Example of cable calculation based on voltage drop

The wire resistance is: R = 2(0.0175×20)/2.5 = 0.28 Ohm.
Current strength in the conductor: I = 7000/220 =31.8 A.
Voltage drop across carrier: U pad. = 31.8×0.28 = 8.9 V.
Voltage drop percentage: U% = (8.9/220)×100 = 4.1 %.

The carrier is suitable for the welding machine in accordance with the requirements of the operating rules for electrical installations, since the percentage of voltage drop across it is within the normal range. However, its value on the supply wire remains large, which can negatively affect the welding process. Here it is necessary to check the lower permissible limit of the supply voltage for the welding machine.

Conclusion

In order to reliably protect electrical wiring from overheating when the rated current is exceeded for a long time, cable cross-sections are calculated based on long-term permissible currents. The calculation is simplified if a load table for the cable cross-section is used. A more accurate result is obtained if the calculation is made based on the maximum current load. And for stable and long-term operation, an automatic switch is installed in the electrical wiring circuit.

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.

Selecting cable cross-section by current - PUE table, calculations and nuances

The Rules for the Management of Electrical Installations clearly state how much current a city apartment should consume in total, and, therefore, what cable cross-section should be used in it. Its parameters: cross-sectional area 2.5 mm?, diameter 1.8 mm, current load 16 A. Of course, an increase in the number of household appliances changes these indicators, so the advice is to use a copper cable with an area of 4 mm?, with a diameter of 2.26 mm, which will withstand a current load of 25 A.

For a private home, these performance indicators are also acceptable. But it is necessary to take into account the fact that in an apartment or house the electrical circuit is divided into circuits (loops), which will be subject to different loads depending on the power of the consumer. Therefore, you will have to select the cable cross-section according to the current (the PUE table is a good helper in this case).

Calculation of wire cross-section

Let's start not with a table, but with a calculation. That is, each person, without having the Internet at hand, where the PUE with tables is freely available, can independently calculate the cable cross-section by current. To do this you will need a caliper and a formula.

If we consider the cross-section of the cable, it is a circle with a certain diameter. There is a formula for the area of a circle:

S= 3.14*D?/4, where 3.14 is the Archimedean number, “D” is the diameter of the measured core. The formula can be simplified: S=0.785*D?.

If the wire consists of several cores, then the diameter of each is measured, the area is calculated, then all the indicators are summed up. How to calculate the cross-section of a cable if each core consists of several thin wires? The process becomes a little more complicated, but not much. To do this, you will have to count the number of wires in one core, measure the diameter of one wire, calculate its area using the described formula and multiply this figure by the number of wires. This will be the cross section of one core. Now you need to multiply this value by the number of cores.

If you don’t want to count the wires and measure their sizes, you just need to measure the diameter of one core, consisting of several wires. You must take measurements carefully so as not to crush the core. Please note that this diameter is not exact because there is space between the wires. Therefore, the resulting value must be multiplied by a reducing factor of 0.91.

Relation between current and cross section

To understand how an electrical cable works, you need to remember a regular water pipe. The larger its diameter, the more water will pass through it. It's the same with wires. The larger their area, the greater the current that will pass through them. In this case, the cable will not overheat, which is the most important requirement of fire safety rules.

Therefore, the cross-section - current connection is the main criterion that is used in the selection of electrical wires in the wiring. Therefore, you need to first figure out how many household appliances and what total power will be connected to each loop. For example, the kitchen must have a refrigerator, microwave, coffee grinder and coffee maker, an electric kettle and sometimes a dishwasher. That is, all these devices can be turned on simultaneously at the same time. Therefore, the total power of the room is used in the calculations.

You can find out the power consumption of each device from the product passport or on the tag. For example, let's outline some of them:

Kettle – 1-2 kW.
Microwave and meat grinder 1.5-2.2 kW.
Coffee grinder and coffee maker – 0.5-1.5 kW.
Refrigerator 0.8 kW.

Having found out the power that will act on the wiring, you can select its cross-section from the table. We will not consider all the indicators in this table; we will show those that prevail in everyday life.

Current strength is 16 A, cable cross-section is 2.7 mm?, wire diameter is 1.87 mm.
25 A – 4.2 – 2.32.
32 A – 5.3 – 2.6.
40 A – 6.7 – 2.92.

But there are nuances here. For example, you need to connect a washing machine. Experts recommend connecting such powerful devices from the switchboard to a separate circuit, powering it to a separate circuit breaker. So the power consumption of the washing machine is 4 kW, and this is a current of 18 A. This indicator is not in the PUE table, so it is necessary to bring it to the nearest larger one, and this is 20 A, to which a circuit with a cross section of 3.3 mm is suitable? diameter 2.05 mm. Again, there is no wire with such a value, which means we bring it to the nearest larger one. Is it 4 mm? By the way, a table of standard sizes of electrical wires is also freely available on the Internet.

Attention! If you don’t have a cable of the required cross-section at hand, you can replace it with two, three, and so on with wires of a smaller area that are connected in parallel. In this case, their total cross-section must coincide with the nominal cross-section. For example, to replace a cable with a cross-section of 10 mm, you can use instead two wires of 5 mm, or three of 2, 3 and 5 mm, or four: two of 2 and two of 3.

Three-phase connection

A three-phase network consists of three wires through which current flows. Accordingly, the load of a device connected to three phases is reduced by three times in each phase. Therefore, a cable with a smaller cross-section can be used for each phase. Here too the ratio is three times. That is, if the cable cross-section in a single-phase network is 4 mm?, then for a three-phase network you can take 4/1.75 = 2.3 mm?. We convert to a standard larger size according to the PUE table - 2.5 mm?.

Aluminum wire

A fairly large number of houses and apartments still have electrical wiring with aluminum cable. There is nothing bad to say about him. Aluminum cable serves well, and as life has shown, its service life is practically unlimited. Of course, if you choose the right current and make the connection correctly.

Just as in the case of a copper cable, we will compare the aluminum one in cross-section, current strength and power. Again, we will not consider everything, we will take only the running parameters.

2.5mm cable? withstands a current of 16 A and a consumer power of 3.5 kW.
4 mm? - 21 A – 4.6 kW.
6 – 26 – 5,7.
10 – 38 – 8,4.

Wire selection

It is best to make internal wiring using copper wires. Although aluminum ones are not inferior to them. But there is one nuance here, which is associated with the correct connection of sections in the distribution box. As practice shows, connections often fail due to oxidation of the aluminum wire.

Another question is which wire to choose: single-core or stranded? Single-core has better current conductivity, so it is recommended for use in household electrical wiring. Multicore has high flexibility, which allows it to be bent in one place several times without compromising quality.

Selecting cable by brand. The best option here is VVG cable. These are copper wires with double plastic insulation. If you come across the “NYM” brand, then consider that it is still the same VVG, only made abroad.

Single and multi-core cable

Attention! It is prohibited to use PUNP brand wires today. For this purpose, there is a decree of Glavgosenergonadzor, which has been in force since 1990.

fix-builder.ru

How to determine the wire cross-section? | Electrician's Notes

Hello, dear readers and guests of the Electrician's Notes website.

When replacing electrical wiring in an apartment with your own hands, many people have a question: “How to determine the cross-section of a wire or cable?”

Most often, citizens are interested in the cross-section of wires or cables that need to be laid from the floor (entrance) to the apartment electrical panel, or from the overhead line support to the input switchgear (IDU) of a cottage or house. No less often, I am asked questions about determining the cross-section of wires and cables for group loads or three-phase motors.

In fact, the issue of choosing the cross-section of wires and cables is very serious, because if the cross-section is insufficient, there will be a high current density in the conductor, and the wire will begin to heat up, thereby destroying the insulation of the wire. Here is an example of the wrong choice of cable cross-section for a socket. Look what this led to.

If we want to use a wire with a larger cross-section, then we need to choose it rationally.

To determine the cross-section of a wire or cable, we will use the PUE tables (Table 1.3.4 - 1.3.11), which indicate long-term permissible currents for copper and aluminum wires (cables, cords) with various types of insulation (PVC, rubber) and sheaths (PVC, lead, nayrite, rubber).

Especially for you, from the PUE tables listed above, I have created one general table from which you can easily determine the cross-section of three-core, four-core and five-core wires and cables for single-phase (220 V) and three-phase (380 V) loads. You only need to know the load current or its power.

Note: in this table the power is calculated at cosφ = 1.

I did not stop the cords, because... When installing and replacing electrical wiring, they are rarely used. You will find long-term permissible currents for SIP wires in GOST 31946-2012 (cancelled GOST R 52373-2005), table 10.

By the way, I would like to take this opportunity to remind you that it is prohibited to use PUNP and APUNP wires (follow the link and read the whole truth about them). Examples of inconsistencies between these wires and the declared cross-section are given not only by me, but also by site visitors.

The rating of the input circuit breaker must be agreed upon by the energy supply organization. It is prohibited to change its denomination on your own, because this affects the selectivity of operation of protection devices installed in the power circuit in an ASU or TP, as well as the allocated power for a specific apartment or house.

The rating of the input circuit breaker can be found in the energy supply organization or in the issued technical conditions (TU) for connection to networks.

Let us assume that, according to the technical specifications, the allocated power for a private house is 5 (kW) single-phase power supply 220 (V), and the rating of the input circuit breaker should be 25 (A).

How to use my table?

Everything is very simple. Depending on the type of electrical wiring (in the air or on the ground), the material of the cores and the voltage, we select the cross-section so that the long-term permissible current of the cable exceeds the rating of the input circuit breaker.

We plan to make the input cable into the house with three-core copper VVGng brand and lay it openly. It turns out that its cross-section should be at least 4 sq. mm, i.e. you need to purchase a VVGng cable (3x4).

But here I recommend remembering such concepts as the “conditional shutdown current” of the machine. Read more about this in the article about the time-current characteristics of machines. It turns out that a machine with a rated current of 25 (A) has a “conditional shutdown current” of 1.45 25 = 36.25 (A). With this current, the machine will turn off in a cold state in about 60 minutes (1 hour). This means that when choosing the cross-section of the power cable, this must be taken into account.

In my example, a cable with a cross section of 4 sq. mm has a continuous permissible current of 35 (A), and the “conditional shutdown current” is 36.25 (A). In principle, the difference between them is small - you can leave it like that. But I would recommend using a 6 sq. mm input cable, which has a long-term permissible current of 42 (A).

How to determine the cross-section of a cable or wire for outlet lines?

Each electrical device has its own installed power and it is indicated in the passport or on a sticker. The unit of measurement is Watt (W).

Let's assume that we need to select a power line for a washing machine whose power is 2.4 (kW). We plan to make the cable with three-core copper VVGng brand and lay it hidden. It turns out that its cross-section should be at least 1.5 sq. mm, i.e. you need to purchase a VVGng cable (3x1.5).

If only the washing machine is connected to this outlet, then the selected VVGng cable (3x1.5) can be left. This cable must be protected with a machine with a rated current of 10 (A).

But I believe that it is not practical to use the outlet for only one washing machine. Surely, you will want to include a hair dryer, electric razor or iron. Therefore, for all outlet lines, I recommend laying a copper cable with a cross-section of 2.5 sq.m., and protecting the line with a machine with a rating of 16 (A).

How to determine the cross-section of a wire (cable) for a three-phase motor?

Let's look at another example. Let's say that at our dacha we have a three-phase asynchronous motor of type AIR71A4U2 with a power of 550 (W), the windings of which are connected by a star to a voltage of 380 (V). We need to select and determine the cross-section of the power cable for it.

We look at the rated current of the motor when connected by a star, indicated on the tag. It is 1.6 (A).

If there is no tag on the motor housing, then the data can be found in the reference tables.

We are planning to purchase a copper power cable and will lay it over the air. We look for the corresponding rows in my table and find the required section.

We get 1.5 sq. mm.

The cross-section of the power cable for the engine can also be determined by its power. Everything is the same.

In the article calculating the cross-section of a cable (wire), I described in detail how to calculate the cross-section using the Electrician program. I also recommend that you read the article on how to determine the cable cross-section by diameter.

After determining the cross-section, you need to proceed to choosing the brand of wires and cables.

P.S. I hope I have presented the material clearly to you and now you can independently determine the cross-section of the wire or cable.

zametkielectrika.ru

Section selection tables

This form can be freely used offline "as is" - i.e. without changing the source text. Regarding the use of the program on websites, you must contact the author - Leonid Miroshko:

Sincerely, Miroshko Leonid.

Tables of PUE and GOST 16442-80 for the WireSel program - Selection of wire cross-section for heating and voltage loss.

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

	open (in tray)	1 + 1(two 1g)	1 + 1 + 1(three 1g)	1 + 1 + 1 + 1(four 1g)	1*2(one 2w)	1*3(one 3zh)
0,5	11	-	-	-	-	-
0,75	15	-	-	-	-	-
1,00	17	16	15	14	15	14
1,5	23	19	17	16	18	15
2,5	30	27	25	25	25	21
4,0	41	38	35	30	32	27
6,0	50	46	42	40	40	34
10,0	80	70	60	50	55	50
16,0	100	85	80	75	80	70
25,0	140	115	100	90	100	85
35,0	170	135	125	115	125	100
50,0	215	185	170	150	160	135
70,0	270	225	210	185	195	175
95,0	330	275	255	225	245	215
120,0	385	315	290	260	295	250
150,0	440	360	330	-	-	-
185,0	510	-	-	-	-	-
240,0	605	-	-	-	-	-
300,0	695	-	-	-	-	-
400,0	830	-	-	-	-	-
	open (in tray)	1 + 1(two 1g)	1 + 1 + 1(three 1g)	1 + 1 + 1 + 1(four 1g)	1 * 2(one 2g)	1 * 3(one 3zh)
	Current loads A of wires laid in one pipe (box, bundle)

PUE, Table 1.3.5. Permissible continuous current for rubber and polyvinyl chloride insulated wires with aluminum conductors
Conductor cross-section, mm2	Current loads A of wires laid in one pipe (box, bundle)
Conductor cross-section, mm2	open (in tray)	1 + 1(two 1g)	1 + 1 + 1(three 1g)	1 + 1 + 1 + 1(four 1g)	1*2(one 2w)	1*3(one 3zh)
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	-	-	-	-	-
Conductor cross-section, mm2	open (in tray)	1 + 1(two 1g)	1 + 1 + 1(three 1g)	1 + 1 + 1 + 1(four 1g)	1 * 2(one 2g)	1 * 3(one 3zh)
Conductor cross-section, mm2	Current loads A of wires laid in one pipe (box, bundle)

PUE, 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
	single-core	two-wire		three-wire
	when laying
	in the air	in the air	in the ground	in the air	in the ground
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	-	-	-	-

PUE, 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 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
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	-	-	-	-

PUE, 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 wires and cables
Conductor cross-section, mm2	single-core	two-wire	three-wire
0.5	-	12	-
0.75	-	16	14
1	-	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

GOST 16442-80, Table 23. Permissible current loads of cables up to 3KV inclusive. with copper conductors with insulation made of polyethylene and polyvinyl chloride plastic, A*
Conductor cross-section, mm2	Current *, A, for wires and cables
	single-core		two-wire		three-wire
	when laying
	in the air	in the ground	in the air	in the ground	in the air	in the ground
1,5	29	32	24	33	21	28
2,5	40	42	33	44	28	37
4	53	54	44	56	37	48
6	67	67	56	71	49	58
10	91	89	76	94	66	77
16	121	116	101	123	87	100
25	160	148	134	157	115	130
35	197	178	166	190	141	158
50	247	217	208	230	177	192
70	318	265	-	-	226	237
95	386	314	-	-	274	280
120	450	358	-	-	321	321
150	521	406	-	-	370	363
185	594	455	-	-	421	406
240	704	525	-	-	499	468

GOST 16442-80, Table 24. Permissible current loads of cables up to 3KV inclusive. with aluminum conductors with insulation made of polyethylene and polyvinyl chloride plastic, A*
Conductor cross-section, mm2	Current *, A, for wires and cables
	single-core		two-wire		three-wire
	when laying
	in the air	in the ground	in the air	in the ground	in the air	in the ground
2.5	30	32	25	33	51	28
4	40	41	34	43	29	37
6	51	52	43	54	37	44
10	69	68	58	72	50	59
16	93	83	77	94	67	77
25	122	113	103	120	88	100
35	151	136	127	145	106	121
50	189	166	159	176	136	147
70	233	200	-	-	167	178
95	284	237	-	-	204	212
120	330	269	-	-	236	241
150	380	305	-	-	273	278
185	436	343	-	-	313	308
240	515	396	-	-	369	355

* Currents apply to wires and cables both with and without a neutral core.

Current loads for wires laid in trays (not in bundles) are the same as for wires laid openly.

To facilitate the selection of cross-section and take into account additional conditions, you can use the form “Calculation of wire cross-section based on permissible heating and permissible voltage loss.” Current values for small cross-sections for copper conductors were obtained by extraplication.

Calculations based on the economic criterion for end consumers are not made.

www.investkabel.ru

Calculation and correct selection of wire and cable cross-sections.

When replacing existing wiring, as well as when laying a new cable or wire, a significant role is played by the correct calculation of the conductor cross-section. After all, oddly enough, how long the electrical wiring will last depends on this.

The first step is to determine what metal the cable or wire is needed from. Aluminum wires have only one plus - low price, but there are a whole lot of minuses. In addition, in the latest versions of the PUE (Electrical Installation Rules) in paragraph 7.1.34 it is written in black and white - “Cables and wires with copper conductors should be used in buildings” and nothing else. But nothing is written about what to do for those who have aluminum wiring, a legacy of a long-defunct country.

If you need to change all the electrical wiring, then there is no problem, we take copper and sleep peacefully. What if you need to change the wiring in only one room and hook it up to the old aluminum one? Then we make a calculation for an aluminum wire and lay it, or we make a calculation for a copper wire and connect it to aluminum through terminals. In no case should you twist it, otherwise you will think for a long time why your apartment burned down (aluminum and copper form a galvanic couple and the place of their direct contact becomes very hot).

The second step is to calculate how many watts the room will consume. To do this, we sum up the power of all electrical appliances that will be in use.

For example: in our room we will have a TV (power 100W), a computer (power 400W), an air conditioner (power 1000W), light (6 bulbs of 60W each), and let’s say a heater (power 2000W). All powers taken as an example are fictitious.

Let's sum up all the powers: 100W + 400W + 1000W + 360W + 2000W = 3860W

The third step is to calculate the current using the formula I=P/U cosФ

I - current (A)P - total power (W)U - network voltage (V)

cosФ (cosine phi) is best taken equal to 1 (if you do not have industrial units)

The network voltage is 220 volts.

We calculate the current strength for our example: I=3860/220·1=17.5 A

Using the table, select the value of the cross-section of the wire or cable (PUE table 1.3.4 and 1.3.5).

Permissible continuous current for wires and cables with copper conductors

Conductor cross-section, mm2
	in one pipe
	two single-core	three single-core	four single-core	one two-wire	one three-wire

Permissible continuous current for wires and cables with aluminum conductors

Conductor cross-section, mm2	Current, A, for wires laid
		in one pipe
		two single-core	three single-core	four single-core	one two-wire	one three-wire

In our case, we use a two-core wire with copper conductors laid in a groove. We select the cross section according to table 1 and it is equal to 1.5 mm2 (with a current of 18 A).

Calculate the wire resistance: R=p·L/S

R - wire resistance (Ohm)p - resistivity (Ohm mm2/m)L - wire or cable length (m)S - cross-sectional area (mm2)

We measure the length of the required wire, take the resistivity from the table and calculate the resistance of the wire or cable.

In our example, we use copper and a wire length of 10 meters.

Substitute the values into the formula: R=0.0175·10/1.5=0.116 Ohm

We calculated the resistance for one core. But since we have a two-core wire, the resistance will be twice as high.

If the wire is three-core, then we also multiply the resistance by 2, only 2 wires are involved, the third is ground.

And the last step is to calculate the voltage loss along the length of the wire. The permissible voltage drop is no more than 5%.

Voltage drop formula: dU=I R

I - current R - wire or cable resistance

dU=17.5·0.232=4.06 V

Converting to percentages: 220 volts is 100%, hence 1% = 2.2 V

dU=4.06/2.2=1.84%

The voltage drop is within acceptable limits, which means the taken cross-section fits perfectly to the given length of the wire. If the voltage drop is more than 5%, then you need to take a larger cross section in the calculations.

To check, we use an online calculation of the cable or wire cross-section.

P.S. I don’t recommend simply calculating the cross-section on an online calculator; it’s good to use it only in conjunction with your calculations, so you definitely won’t make a mistake and choose the correct cross-section of the wire or cable.

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Continuous-permissible cable current according to PUE - table and explanations

Currents flowing through the cable heat the conductor. This does not apply to the beneficial effects of current, such as heating the filament of a light bulb or an electric stove. Therefore, we do not take this action into account when calculating the total power consumption. However, you should not forget about the energy consumption for heating the wires, as this can lead to dire consequences.

The amount of current flowing through the wires depends on the power of the consumer devices, since the power released on the wires themselves is negligible - due to the low resistivity of the metals used for the wire and in the wiring cable. Current flows only when we plug in devices. In this case, the total current at each moment in time is determined only by the power of the devices (related to the resistance) consuming energy in the network at that particular moment in time. But when calculating the network by current and power, it is always necessary to take only situations when all consuming devices are turned on at the same time. Only this approach makes it possible to insure against all possible overloads. But that's not all. At the moment of switching on, many devices consume the so-called starting current, which can be 10–20 percent higher in consumption from the stationary operation of the device. For some devices, this is due to the difficulty of starting - accelerating massive rotors, creating operating pressure drops, and so on. Therefore, when performing the calculation, it is also necessary to make an adjustment for this.

Permissible continuous current for cables

Current-carrying wires always heat up when exposed to current. The only question is the amount of heat generated. On the one hand, it depends on the flowing current, the resistivity of the conductor material, its cross-section, on the other hand, on the factors of heat removal in the conditions of the passage of wires: on the number of wires and their proximity, insulation, which prevents heat removal, the presence of boxes or channels into which The cable is tucked in, the wiring is hidden. And in general, from the climatic factors acting on the cable in the places where the wires pass: ventilation, open space, and so on.

Wiring quality and aging

As a result of all these numerous factors, a wire that is systematically heated by the current passing through it, from a safety point of view, can be:

Reliable carrier of current and voltage. For such a wire, the period of future trouble-free operation can be considered unlimited.
Old or aging electrical energy carrier. The quality of the wire has decreased during operation, the insulation has deteriorated, the joints and connections of the wires have lost some of their conductivity. Wire aging tends to accumulate over time and contribute to an increase in the rate of aging and an increase in negative factors.
Dangerous electrical wiring. The operating mode is such that accidents are likely. This is expressed in increased heating of wires at normal current, uneven heating due to deterioration of insulation, oxidation of contacts, deterioration in the uniformity of wire cross-section due to oxidation, which is natural for metals. Unevenness also tends to enhance aging and locally degrade quality.

Temperature is therefore a very important indicator of the safety of electrical wiring. In addition, the temperature regime itself can deteriorate the wiring, and if the maximum threshold is exceeded, lead to accidents. As a result, the permissible current loads of the cables must be reduced.

For example, there is a rule that every 8° of excess current heating of the cable speeds up the processes (both chemical and physical) in the material by a factor of two. This affects the performance of the conductor (especially aluminum) and degrades the performance of the insulator.

Insulation and temperature

Insulation as a result of heating can itself become a source of dangerous and harmful factors. For example, PVC behaves like this when the temperature increases:

80 °C - softening;
100 °C - release of HCl (volatile harmful gas, hydrogen chloride, which when dissolved in water becomes hydrochloric acid). As the temperature rises, the process intensifies. At 160 °C 50% of it will already be released, at 300 °C - 85%;
210 °C - melting;
350 °C - the PVC carbon base begins to ignite.

This applies to hard PVC; soft PVC contains many plasticizer additives that volatilize and can catch fire even at 200 °C.

Softening, especially melting, conceals another danger - current-carrying wires may come closer together, which usually leads to a short circuit and fire.

For safety reasons, the upper temperature limit of the wires through which electric current passes was set at 65 °C. This is at an ambient air temperature of 25 °C, ground temperature - 15 °C.

The task of maintaining such a heating rate is to select, for all the variety of conditions, cross-sections for wires made of different materials used in electrical engineering that are sufficient for the safe passage of current, that is, without heat accumulation.

A prerequisite is that the permissible long-term current for the cables is meant, and not short-term overloads.

Wires and cables should be protected from sudden overcurrents by circuit breakers on the power panel.

Moreover, their ratings are selected so that they are higher than the currents that arise during short-term, but permissible overloads, but lower than the overvoltages that are dangerous for the network.

Wiring structure of the consuming network

The consumption network consists of several groups of consumers. Each of them has its own nature of loads and current regime, therefore, the wiring must comply with safety rules. The most important rule: high loadability must be ensured where it is loaded. That is, the input wires that bear the entire burden of consumption in the network must be the largest in cross-section, since through them the energy is consumed for the entire power of the loads in the network in question.

Example. Calculation of cable cross-section for residential consuming network

The table shows consumption devices

Bus current from the total power formula

with KI, utilization factor equal to 75% and cos j = 1,

is obtained in the range I = 41–81 A. For wiring that takes into account any possible options for the power of connected electrical appliances, you should take the upper value and a margin for the future of about 10–20%. Therefore, we accept a maximum current of 100 A.

Perhaps such a load will place a heavy burden on the buses of the home network, and the electricity supply organization will not allow so many consumers at once, but the choice of wires should not depend on such “political” issues. Moreover, wiring in old houses already demonstrates the short-sightedness of previous restrictions.

The cross-section of the buses supplied to the apartments must be taken as a given. If we do the wiring in the apartment ourselves, then we divide it into several subnetworks according to groups of current consuming devices. Each subnetwork will be powered separately from the power panel buses. And it must be performed with the expectation of maximum consumption in this particular subnet.

PUE - rules for electrical installations

To regulate safety relating to everything related to electricity, there is a system of rules that began to be developed from the very beginning of the use of electricity (1899, the First All-Russian Electrotechnical Congress) and were brought into a system close to the modern one, immediately after the Great Patriotic War in 1946– 1949 And they exist and continue to be developed now - in Russia, Belarus and Ukraine.

Electrical safety is a very serious matter, despite differences of opinion elsewhere. For example, we also provide for fines for non-compliance with the rules for the installation of electrical installations for citizens, officials and entrepreneurs and for legal entities.

What concerns electrical wiring safety is collected in section 1 in chapter 3.

The tables show the permissible continuous current for cables for a variety of wire options, metals (different resistivities), insulation, nature (single-core - stranded), wire cross-section, as well as cable laying methods.

The full text of Chapter 3 from Section 1 of the 7th edition of the PUE is available in the following file. The permissible continuous current for cables in them is presented in tables 3.1.7.4 – 3.1.7.11.

For our example, we will build a table, dividing all consumers into groups, in each group we will calculate the total power and current and find from the PUE the corresponding cable cross-section for copper and aluminum.

In our case, we will select subnetworks and calculate the total power and maximum current for each of them. From the PUE we will select the wire cross-section for copper wires and aluminum:

It turns out that a wire cross-section of 1 mm2 copper or 2 mm2 aluminum is suitable for the lighting network.

For an outlet network with low consumption (residential premises), 1.5 and 2.5 mm2, respectively.

Two outlet subnetworks with a significant level of consumption - in the kitchen and bathroom - gave 4 and 5–6 mm2.

Individual consumers can be powered by separate wiring with individual calculation of current and cross-section.