Cable cross-section for 20 kW three phases. Calculation of cable cross-section by power: practical advice from professionals. Calculation by length

In order to correctly install electrical wiring, ensure uninterrupted operation of the entire electrical system and eliminate the risk of fire, before purchasing a cable, it is necessary to calculate the loads on the cable to determine the required cross-section.

There are several types of loads, and for the highest quality installation of the electrical system, it is necessary to calculate the loads on the cable according to all indicators. The cable cross-section is determined by load, power, current and voltage.

Power section calculation

In order to produce, it is necessary to add up all the indicators of electrical equipment operating in the apartment. Calculation of electrical loads on the cable is carried out only after this operation.

Calculation of cable cross-section by voltage

Calculation of electrical loads on a wire necessarily includes. There are several types electrical network- single-phase for 220 volts, as well as three-phase - for 380 volts. In apartments and residential premises, as a rule, a single-phase network is used, so in the calculation process it is necessary to take into account at the moment— in the tables for calculating the cross-section, the voltage must be indicated.

Calculation of cable cross-section by load

Table 1. Installed power(kW) for cables laid openly

Core cross-section, mm 2	Cables with copper conductors		Cables with aluminum conductors
	220 V	380 V	220 V	380 V
0,5	2,4
0,75	3,3
1	3,7	6,4
1,5	5	8,7
2	5,7	9,8	4,6	7,9
2,5	6,6	11	5,2	9,1
4	9	15	7	12
5	11	19	8,5	14
10	17	30	13	22
16	22	38	16	28
25	30	53	23	39
35	37	64	28	49

Table 2. Installed power (kW) for cables laid in a groove or pipe

Core cross-section, mm 2	Cables with copper conductors		Cables with aluminum conductors
	220 V	380 V	220 V	380 V
0,5
0,75
1	3	5,3
1,5	3,3	5,7
2	4,1	7,2	3	5,3
2,5	4,6	7,9	3,5	6
4	5,9	10	4,6	7,9
5	7,4	12	5,7	9,8
10	11	19	8,3	14
16	17	30	12	20
25	22	38	14	24
35	29	51	16

Each electrical appliance installed in the house has a certain power - this indicator indicated on the nameplates of devices or in technical passport equipment. To implement, it is necessary to calculate the total power. When calculating the cable cross-section for the load, it is necessary to rewrite all electrical equipment, and you also need to think about what equipment may be added in the future. Since installation is carried out on long term, it is necessary to take care of this issue so that a sharp increase in load does not lead to an emergency.

For example, you have a total voltage of 15,000 W. Since the vast majority of residential premises have a voltage of 220 V, we will calculate the power supply system taking into account a single-phase load.

Next, you need to consider how much equipment can operate simultaneously. As a result, you will get a significant figure: 15,000 (W) x 0.7 (70% simultaneity factor) = 10,500 W (or 10.5 kW) - the cable must be designed for this load.

You also need to determine what material the cable cores will be made of, since different metals have different conductive properties. Mainly used in residential areas copper cable, since its conductive properties far exceed those of aluminum.

It is worth considering that the cable must have three cores, since grounding is required for the electrical supply system in the premises. In addition, it is necessary to determine what type of installation you will use - open or hidden (under plaster or in pipes), since the calculation of the cable cross-section also depends on this. Once you have decided on the load, core material and type of installation, you can look at the required cable cross-section in the table.

Calculation of cable cross-section for current

First you need to calculate the electrical loads on the cable and find out the power. Let's say that the power turned out to be 4.75 kW, we decided to use a copper cable (wire) and lay it in a cable channel. is produced according to the formula I = W/U, where W is power, and U is voltage, which is 220 V. In accordance with this formula, 4750/220 = 21.6 A. Next, look at table 3, we get 2, 5 mm.

Table 3. Permissible current loads for cables with copper conductors laid hidden

Core cross-section, mm	Copper conductors, wires and cables
	Voltage 220 V	Voltage 380 V
1,5	19	16
2,5	27	25
4	38	30
6	46	40
10	70	50
16	85	75
25	115	90
35	135	115
50	175	145
70	215	180
95	260	220
120	300	260

Hello!

I have heard about some difficulties that arise when choosing equipment and connecting it (what outlet is needed for the oven, hob or washing machine). In order for you to quickly and easily solve this, as good advice, I suggest you familiarize yourself with the tables presented below.

Types of equipment	Included	What else is needed
	terminals
Email panel (independent)	terminals	cable supplied from the machine, with a margin of at least 1 meter (for connection to the terminals)
		euro socket
Gas panel		gas hose, euro socket
Gas oven	cable and plug for electric ignition	gas hose, euro socket
Washing machine
Dishwasher	cable, plug, hoses about 1300mm. (drain, bay)	for connection to water, ¾ outlet or straight-through tap, Euro socket
Refrigerator, wine cabinet	cable, plug	euro socket
Hood	cable, plug may not be included	corrugated pipe(at least 1 meter) or PVC box, Euro socket
Coffee machine, steamer, microwave oven	cable, plug	euro socket

Types of equipment		Socket	Cable cross-section		Automatic + RCD⃰ in the panel
			Single-phase connection	Three-phase connection
Dependent set: el. panel, oven	about 11 kW (9)		6mm² (PVS 3*6) (32-42)	4mm² (PVS 5*4) (25)*3	separate at least 25A (380V only)
Email panel (independent)	6-15 kW (7)		up to 9 kW/4mm² 9-11 kW/6mm² 11-15KW/10mm² (PVS 4,6,10*3)	up to 15 kW/ 4mm² (PVS 4*5)	separate at least 25A
Email oven (independent)	about 3.5 - 6 kW	euro socket	2.5mm²		not less than 16A
Gas panel		euro socket	1.5mm²		16A
Gas oven		euro socket	1.5mm²		16A
Washing machine	2.5 kW	euro socket	2.5mm²		separate at least 16A
Dishwasher	2 kW	euro socket	2.5mm²		separate at least 16A
Refrigerator, wine cabinet	less than 1KW	euro socket	1.5mm²		16A
Hood	less than 1KW	euro socket	1.5mm²		16A
Coffee machine, steamer	up to 2 kW	euro socket	1.5mm²		16A

⃰ Residual current device

Electrical connection at voltage 220V/380V

Types of equipment	Maximum power consumption	Socket	Cable cross-section		Automatic + RCD⃰ in the panel
			Single-phase connection	Three-phase connection
Dependent set: el. panel, oven	about 9.5KW	Calculated for the power consumption of the kit	6mm² (PVS 3*3-4) (32-42)	4mm² (PVS 5*2.5-3) (25)*3	separate at least 25A (380V only)
Email panel (independent)	7-8 kW (7)	Calculated for panel power consumption	up to 8 kW/3.5-4mm² (PVS 3*3-4)	up to 15 kW/ 4mm² (PVS 5*2-2.5)	separate at least 25A
Email oven (independent)	about 2-3 kW	euro socket	2-2.5mm²		not less than 16A
Gas panel		euro socket	0.75-1.5mm²		16A
Gas oven		euro socket	0.75-1.5mm²		16A
Washing machine	2.5-7(with drying) kW	euro socket	1.5-2.5mm²(3-4mm²)		separate at least 16A-(32)
Dishwasher	2 kW	euro socket	1.5-2.5mm²		separate at least 10-16A
Refrigerator, wine cabinet	less than 1KW	euro socket	1.5mm²		16A
Hood	less than 1KW	euro socket	0.75-1.5mm²		6-16A
Coffee machine, steamer	up to 2 kW	euro socket	1.5-2.5mm²		16A

When choosing a wire, first of all you should pay attention to the rated voltage, which should not be less than in the network. Secondly, you should pay attention to the material of the cores. Copper wire has greater flexibility than aluminum wire and can be soldered. Aluminum wires Do not lay over combustible materials.

You should also pay attention to the cross-section of the conductors, which must correspond to the load in amperes. You can determine the current in amperes by dividing the power (in watts) of all connected devices by the voltage in the network. For example, the power of all devices is 4.5 kW, voltage 220 V, which is 24.5 amperes. Use the table to find the required cable cross-section. It will be copper wire with a cross section of 2 mm 2 or aluminum wire with a cross section of 3 mm 2. When choosing a wire of the cross-section you need, consider whether it will be easy to connect to electrical devices. The wire insulation must correspond to the installation conditions.

Laid open
S	Copper conductors			Aluminum conductors
mm 2	Current	Power kW		Current	Power kW
	A	220 V	380 V	A	220 V	380 V
0,5	11	2,4
0,75	15	3,3
1	17	3,7	6,4
1,5	23	5	8,7
2	26	5,7	9,8	21	4,6	7,9
2,5	30	6,6	11	24	5,2	9,1
4	41	9	15	32	7	12
6	50	11	19	39	8,5	14
10	80	17	30	60	13	22
16	100	22	38	75	16	28
25	140	30	53	105	23	39
35	170	37	64	130	28	49

Installed in a pipe
S	Copper conductors			Aluminum conductors
mm 2	Current	Power kW		Current	Power kW
	A	220 V	380 V	A	220 V	380 V
0,5
0,75
1	14	3	5,3
1,5	15	3,3	5,7
2	19	4,1	7,2	14	3	5,3
2,5	21	4,6	7,9	16	3,5	6
4	27	5,9	10	21	4,6	7,9
6	34	7,4	12	26	5,7	9,8
10	50	11	19	38	8,3	14
16	80	17	30	55	12	20
25	100	22	38	65	14	24
35	135	29	51	75	16	28

Wire markings.

The 1st letter characterizes the material of the conductor:
aluminum - A, copper - the letter is omitted.

The 2nd letter means:
P - wire.

The 3rd letter indicates the insulation material:
B - shell made of polyvinyl chloride plastic,
P - polyethylene shell,
R - rubber shell,
N—nairite shell.
Marks of wires and cords may also contain letters characterizing other structural elements:
O - braid,
T - for installation in pipes,
P - flat,
F-t metal folded shell,
G - increased flexibility,
And - increased protective properties,
P - braided cotton yarn impregnated with an anti-rotten compound, etc.
For example: PV - copper wire with polyvinyl chloride insulation.

Installation wires PV-1, PV-3, PV-4 are intended for supplying power to electrical devices and equipment, as well as for stationary installation of lighting electrical networks. PV-1 is produced with a single-wire conductive copper conductor, PV-3, PV-4 - with twisted conductors from copper wire. The wire cross-section is 0.5-10 mm 2. The wires have painted PVC insulation. Used in AC circuits rated voltage no more than 450 V with a frequency of 400 Hz and in circuits DC with voltage up to 1000 V. Operating temperature limited to the range -50…+70 °C.

Installation PVS wire designed to connect electrical appliances and equipment. The number of cores can be 2, 3, 4 or 5. The conductive core made of soft copper wire has a cross-section of 0.75-2.5 mm 2. Available with twisted conductors in PVC insulation and the same sheath.

It is used in electrical networks with a rated voltage not exceeding 380 V. The wire is designed for a maximum voltage of 4000 V, with a frequency of 50 Hz, applied for 1 minute. Operating temperature - in the range -40...+70 °C.

The PUNP installation wire is intended for laying stationary lighting networks. The number of cores can be 2.3 or 4. The cores have a cross-section of 1.0-6.0 mm 2. The conductor is made of soft copper wire and has plastic insulation in a PVC sheath. It is used in electrical networks with a rated voltage of no more than 250 V with a frequency of 50 Hz. The wire is rated for a maximum voltage of 1500 V at a frequency of 50 Hz for 1 minute.

Power cables of the VVG and VVGng brands are intended for transmission electrical energy in stationary installations AC. The cores are made of soft copper wire. The number of cores can be 1-4. Cross-section of current-carrying conductors: 1.5-35.0 mm 2 . The cables are produced with an insulating sheath made of polyvinyl chloride (PVC) plastic. VVGng cables have reduced flammability. Used with a rated voltage of no more than 660 V and a frequency of 50 Hz.

NYM brand power cable is designed for industrial and household permanent installation indoors and on outdoors. The cable wires have a single-wire copper core with a cross-section of 1.5-4.0 mm 2, insulated with PVC plastic. The outer shell, which does not support combustion, is also made of light gray PVC plastic.

This seems to be the main thing that it is advisable to understand when choosing equipment and wires for them))

The article discusses the main criteria for choosing a cable cross-section and gives examples of calculations.

In markets you can often see handwritten signs indicating which one the buyer needs to purchase depending on the expected load current. Do not believe these signs, as they are misleading. The cable cross-section is selected not only by the operating current, but also by several other parameters.

First of all, it is necessary to take into account that when using a cable at the limit of its capabilities, the cable cores heat up by several tens of degrees. The current values ​​shown in Figure 1 assume heating of the cable cores to 65 degrees at a temperature environment 25 degrees. If several cables are laid in one pipe or tray, then due to their mutual heating (each cable heats all other cables), the maximum permissible current is reduced by 10 - 30 percent.

Also, the maximum possible current decreases when elevated temperature environment. Therefore, in a group network (network from panels to lamps, plug sockets and other electrical receivers) as a rule, cables are used at currents not exceeding 0.6 - 0.7 of the values ​​shown in Figure 1.

Rice. 1. Permissible long-term current of cables with copper conductors

Based on this, the widespread use of circuit breakers with a rated current of 25A to protect socket networks laid with cables with copper conductors with a cross-section of 2.5 mm2 is dangerous. Tables of reduction coefficients depending on temperature and the number of cables in one tray can be found in the Electrical Installation Rules (PUE).

Additional restrictions arise when the cable is longer. In this case, voltage losses in the cable can reach unacceptable values. As a rule, when calculating cables, the maximum loss in the line is no more than 5%. Losses are not difficult to calculate if you know the resistance value of the cable cores and rated current loads. But usually, to calculate losses, they use tables of the dependence of losses on the load moment. The load moment is calculated as the product of the cable length in meters and the power in kilowatts.

Data for calculating losses at a single-phase voltage of 220 V are shown in Table 1. For example, for a cable with copper conductors with a cross-section of 2.5 mm2, with a cable length of 30 meters and a load power of 3 kW, the load moment is 30x3 = 90, and the losses will be 3%. If the calculated loss value exceeds 5%, then it is necessary to select a cable with a larger cross-section.

Table 1. Load moment, kW x m, for copper conductors in a two-wire line for a voltage of 220 V at a given conductor cross-section

Using Table 2, you can determine the losses in a three-phase line. Comparing tables 1 and 2, you can see that in a three-phase line with copper conductors with a cross-section of 2.5 mm2, losses of 3% correspond to six times the load torque.

A triple increase in the load torque occurs due to the distribution of load power across three phases, and a double increase due to the fact that in three-phase network with a symmetrical load (identical currents in the phase conductors), the current in the neutral conductor is zero. With an asymmetrical load, cable losses increase, which must be taken into account when choosing the cable cross-section.

Table 2. Load moment, kW x m, for copper conductors in a three-phase four-wire line with zero for a voltage of 380/220 V at a given conductor cross-section (to enlarge the table, click on the figure)

Cable losses have a significant impact when using low-voltage lamps, such as halogen lamps. This is understandable: if 3 Volts drop on the phase and neutral conductors, then at a voltage of 220 V we most likely will not notice this, and at a voltage of 12 V, the voltage on the lamp will drop by half to 6 V. That is why transformers for powering halogen lamps need to be maximally bring it closer to the lamps. For example, with a cable length of 4.5 meters with a cross-section of 2.5 mm2 and a load of 0.1 kW (two 50 W lamps), the load torque is 0.45, which corresponds to a loss of 5% (Table 3).

Table 3. Load moment, kW x m, for copper conductors in a two-wire line for a voltage of 12 V at a given conductor cross-section

The tables above do not take into account the increase in resistance of conductors due to heating due to current flowing through them. Therefore, if the cable is used at currents of 0.5 or more of the maximum permissible current cable of a given cross-section, then a correction must be made. In the simplest case, if you expect losses of no more than 5%, then calculate the cross section based on losses of 4%. Also, losses may increase if there is large quantity connections of cable cores.

Cables with aluminum conductors have a resistance 1.7 times greater than cables with copper conductors, and accordingly their losses are 1.7 times greater.

The second limiting factor when long lengths cable is exceeding the permissible resistance value of the phase-zero circuit. To protect cables from overloads and short circuits, as a rule, circuit breakers with a combined release are used. Such switches have thermal and electromagnetic releases.

The electromagnetic release provides instantaneous (tenths and even hundredths of a second) shutdown of the emergency section of the network in the event of a short circuit. For example, a circuit breaker designated C25 has a 25 A thermal release and a 250 A electromagnetic release. Automatic circuit breakers of group “C” have a multiplicity of the breaking current of the electromagnetic release to the thermal one from 5 to 10. But the maximum value is taken.

IN total resistance phase-zero circuits are included: the resistance of the step-down transformer of the transformer substation, the resistance of the cable from the substation to the input switchgear of the building, the resistance of the cable laid from the ASU to switchgear(RU) and the cable resistance of the group line itself, the cross-section of which must be determined.

If the line has a large number of connections of cable cores, for example, a group line consisting of a large number of lamps connected by a cable, then the resistance contact connections also subject to accounting. Very accurate calculations take into account the arc resistance at the fault point.

Circuit impedance phase - zero for four-core cables are given in Table 4. The table takes into account the resistance of both the phase and neutral conductors. Resistance values ​​are given at a cable core temperature of 65 degrees. The table is also valid for two-wire lines.

Table 4. Circuit impedance phase - zero for 4-core cables, Ohm/km at core temperature 65 o C

In urban transformer substations, as a rule, transformers with a capacity of 630 kV or more are installed. A and more, having an output resistance Rtp less than 0.1 Ohm. In rural areas, transformers of 160 - 250 kV can be used. And having an output resistance of about 0.15 Ohm, and even transformers for 40 - 100 kV. A, having an output impedance of 0.65 - 0.25 Ohm.

Power supply cables from city transformer substations for ASUs of houses, as a rule, they are used with aluminum conductors with a cross-section of phase conductors of at least 70 - 120 mm2. If the length of these lines is less than 200 meters, the resistance of the phase-neutral circuit of the supply cable (Rpc) can be taken equal to 0.3 Ohm. For a more accurate calculation, you need to know the length and cross-section of the cable, or measure this resistance. One of the devices for such measurements (Vector device) is shown in Fig. 2.

Rice. 2. Device for measuring the resistance of the phase-zero circuit "Vector"

The line resistance must be such that in the event of a short circuit, the current in the circuit is guaranteed to exceed the operating current of the electromagnetic release. Accordingly, for circuit breaker C25 the current short circuit in the line must exceed the value of 1.15x10x25=287 A, here 1.15 is the safety factor. Therefore, the resistance of the phase-zero circuit for the C25 circuit breaker should be no more than 220V/287A=0.76 Ohm. Accordingly, for the C16 circuit breaker the circuit resistance should not exceed 220V/1.15x160A=1.19 Ohms and for the C10 circuit breaker - no more than 220V/1.15x100=1.91 Ohms.

Thus, for urban apartment building, taking Rtp=0.1 Ohm; Rpk=0.3 Ohm when using a cable with copper conductors with a cross section of 2.5 mm2, protected in the socket network circuit breaker C16, cable resistance Rgr (phase and neutral conductors) should not exceed Rgr = 1.19 Ohm - Rtp - Rpk = 1.19 - 0.1 - 0.3 = 0.79 Ohm. From Table 4 we find its length - 0.79/17.46 = 0.045 km, or 45 meters. For most apartments this length is sufficient.

When using a C25 circuit breaker to protect a cable with a cross-section of 2.5 mm2, the circuit resistance must be less than 0.76 - 0.4 = 0.36 Ohm, which corresponds to a maximum cable length of 0.36/17.46 = 0.02 km, or 20 meters.

When using a C10 circuit breaker to protect a group lighting line made with a cable with copper conductors with a cross-section of 1.5 mm2, we obtain the maximum permissible resistance cable 1.91 - 0.4 = 1.51 Ohm, which corresponds to a maximum cable length of 1.51/29.1 = 0.052 km, or 52 meters. If such a line is protected by a C16 circuit breaker, then maximum length line will be 0.79/29.1 = 0.027 km, or 27 meters.

For correct and safe installation of wiring cables, it is imperative to make a preliminary calculation of the expected power consumption. Failure to comply with the requirements for selecting the cross-section of the cable used for wiring can lead to insulation melting and fire.

Calculating the cable cross-section for a specific electrical wiring system can be divided into several stages:

1. breakdown of electricity consumers by groups;
2. definition maximum current for each segment;
3. selection of cable cross-section.

All consuming electrical appliances should be divided into several groups so that the total power consumption of one group does not exceed approximately 2.5-3 kW. This will allow you to select a copper cable with a cross-section of no more than 2.5 square meters. mm. Power of some basic household appliances is shown in Table 1.

Table 1. Power values ​​of major household appliances.

Consumers combined into one group must be geographically located in approximately the same place, since they are connected to the same cable. If the entire connected object is powered from single-phase network, then the number of groups and the distribution of consumers do not play a significant role.

Then the percentage of discrepancy can be calculated using the formula = 100% — (Pmin/Pmax*100%), where Pmax is the maximum total power per phase, Pmin is the minimum total power per phase. The lower the power discrepancy percentage, the better.

Calculation of the maximum current for each consumer group

Once the power consumption has been found for each group, the maximum current can be calculated. It is better to take the demand coefficient (Kc) equal to 1 everywhere, since the use of all elements of one group at the same time is not excluded (for example, you can simultaneously turn on all household appliances belonging to one group of consumers). Then the formulas for single-phase and three-phase networks will look like:

Icalc = Pcalc / (Unom * cosφ)
for a single-phase network, in this case the network voltage is 220 V,

Icalc = Pcalc / (√3 * Unom * cosφ)
for three-phase network, network voltage 380 V.

When installing electrical wiring in recent decades, the method using. This is explained by a whole set of properties that a corrugated pipe has, but at the same time, when working with it, you must adhere to certain rules.

You can often come across both in theory and in practice the terms delta and star connection, phase and linear voltage - an interesting one will help you understand their differences.

The cosine value for household appliances and incandescent lighting is taken equal to 1, for LED lighting– 0.95, for fluorescent lighting– 0.92. The arithmetic mean cosine is found for the group. Its value depends on the cosine of the device consuming highest power in this group. Thus, knowing the currents in all sections of the wiring, you can begin to select the cross-section of wires and cables.

Selection of cable cross-section based on power

Once the calculated maximum current is known, you can begin selecting cables. This can be done in two ways, but the easiest way is to select the desired cable cross-section using the tabular data. The parameters for selecting copper and aluminum cables are given in the table below.

Table 2. Data for selecting the cross-section of a cable with copper conductors and a cable made of aluminum.

When planning electrical wiring, it is preferable to choose cables from the same material. Connection of copper and aluminum wires regular twisting is prohibited by the rules fire safety, since when temperature fluctuates, these metals expand differently, which leads to the formation of gaps between the contacts and the generation of heat. If there is a need to connect cables from different materials, then it is best to use terminals specially designed for this purpose.

Video with formulas for calculating cable cross-section

Proper selection of cables for restoration or installation of electrical wiring guarantees flawless operation of the system. The devices will receive full power. There will be no overheating of the insulation with subsequent destructive consequences. A reasonable calculation of the wire cross-section in terms of power will eliminate both the threat of ignition and extra costs to buy expensive wire. Let's look at the calculation algorithm.

In simple terms, a cable can be compared to a pipeline transporting gas or water. In the same way, a flow moves along its core, the parameters of which are limited by the size of a given current-carrying channel. The consequence of incorrect selection of its cross-section are two common erroneous options:

• The current-carrying channel is too narrow, due to which the current density increases significantly. An increase in current density entails overheating of the insulation, then its melting. As a result of melting, “weak” places for regular leaks will appear at a minimum, and at a maximum there will be a fire.
• The vein is too wide, which is actually not bad at all. Moreover, the presence of space for transporting electric current has a very positive effect on the functionality and operational life of the wiring. However, the owner’s pocket will be lightened by an amount approximately twice the amount actually required.

The first of the erroneous options represents an outright danger, in best case scenario will result in an increase in electricity bills. The second option is not dangerous, but extremely undesirable.

Well-trodden paths of computing

All existing calculation methods rely on Ohm's law, according to which current multiplied by voltage equals power. Household voltage is a constant value, equal to standard 220 V in a single-phase network. This means that only two variables remain in the legendary formula: current and power. You can and should “dance” in calculations from one of them. Using the calculated values ​​of current and expected load in the PUE tables, we will find the required cross-sectional size.

Please note that the cable cross-section is calculated for power lines, i.e. for wires to sockets. Lighting lines are a priori laid with a cable with a traditional cross-sectional area of ​​1.5 mm².

If the room being equipped does not have a powerful disco spotlight or chandelier that requires a power supply of 3.3 kW or more, then it makes no sense to increase the cross-sectional area of ​​the lighting cable core. But the rosette issue is a purely individual matter, because... Such unequal tandems as a hair dryer with a water heater or an electric kettle with a microwave can be connected to the same line.

For those who plan to load power line electrical hob, boiler, washing machine and similar “gluttonous” equipment, it is advisable to distribute the entire load over several outlet groups.

If technical feasibility there is no splitting the load into groups, experienced electricians It is recommended to lay a cable with a copper conductor cross-section of 4-6 mm² without any fuss. Why with a copper current-carrying core? Because the strict PUE code prohibits laying cables with aluminum “filling” in housing and in actively used domestic premises. Electrical copper has much less resistance, it passes more current and does not heat up like aluminum. Aluminum wires are used in the construction of external overhead networks; in some places they still remain in old houses.

Pay attention! The cross-sectional area and diameter of the cable core are two different things. The first is indicated in square mm, the second simply in mm. The main thing is not to confuse!

To search for tabular values ​​of power and permissible current, you can use both indicators. If the table shows the size of the cross-sectional area in mm², and we only know the diameter in mm, the area must be found using the following formula:

Calculation of section size based on load

The simplest way to select a cable with the right size- calculation of wire cross-section according to total power all units connected to the line.

The calculation algorithm is as follows:

• First, let's decide on the units that we can presumably use at the same time. For example, while the boiler is operating, we suddenly want to turn on the coffee grinder, hair dryer and washing machine;
• then, according to data from technical passports or according to approximate information from the table below, we simply sum up the power of household units simultaneously operating according to our plans;
• Let's assume that in total we have 9.2 kW, but this specific value is not in the PUE tables. This means you will have to round to a safe value. big side– i.e. take the closest value with some excess power. This will be 10.1 kW and the corresponding cross-sectional value is 6 mm².

We direct all roundings upward. In principle, it is possible to sum up the current strength indicated in the data sheets. Calculations and rounding for current are carried out in a similar way.

How to calculate the current cross section?

Table values ​​cannot take into account individual characteristics devices and network operation. The specificity of the tables is average. They do not list the parameters of the maximum permissible currents for a particular cable, but they differ for products with different brands. The type of gasket is touched upon very superficially in the tables. To the meticulous masters who reject easy way searching the tables, it is better to use the method of calculating the size of the wire cross-section by current. More precisely, by its density.

Allowable and operating current density

Let's start with mastering the basics: remember in practice the derived interval 6 - 10. These are the values ​​​​obtained by electricians over many years of “experimental methods”. The strength of the current flowing through 1 mm² of copper core varies within the specified limits. Those. a cable with a copper core with a cross-section of 1 mm² without overheating and melting of the insulation allows a current of 6 to 10 A to easily reach the waiting consumer unit. Let's figure out where it came from and what the designated interval fork means.

According to the code electrical laws PUE 40% is allocated to the cable for overheating that is not dangerous for its sheath, which means:

• 6 A distributed per 1 mm² of current-carrying core is the normal operating current density. Under these conditions, the conductor can work indefinitely without any time restrictions;
• 10 A distributed per 1 mm² of copper core can flow through the conductor for a short time. For example, when you turn on the device.

An energy flow of 12 A in a copper millimeter channel will initially be “crowded”. Due to the crowding and crowding of electrons, the current density will increase. The temperature of the copper component will then increase, which will invariably affect the condition of the insulating shell.

Please note that for a cable with aluminum current-carrying conductor, the current density displays an interval of 4 - 6 Amperes per 1 mm² of conductor.

We found out that the maximum current density for a conductor made of electrical copper is 10 A per cross-sectional area of ​​1 mm², and normal is 6 A. Therefore:

• a cable with a conductor cross-section of 2.5 mm² will be able to transport a current of 25 A in just a few tenths of a second when the equipment is turned on;
• it will be able to transmit a current of 15A indefinitely.

The above current densities are valid for open wiring. If the cable is laid in the wall, metal sleeve or, the specified current density value must be multiplied by a correction factor of 0.8. Remember one more subtlety in organization open type wiring. For reasons of mechanical strength, cables with a cross-section of less than 4 mm² in open circuits do not use.

Studying the calculation scheme

There will be no super complex calculations again; calculating the wire for the upcoming load is extremely simple.

• First let's find the limit permissible load. To do this, we summarize the power of the devices that we plan to simultaneously connect to the line. Let's add up, for example, the power of a washing machine 2000 W, a hair dryer 1000 W and an arbitrary heater 1500 W. We received 4500 W or 4.5 kW.
• Then we divide our result by the standard voltage value of a household network of 220 V. We got 20.45 ... A, we round up to a whole number, as expected.
• Next, we introduce a correction factor, if necessary. The value with the coefficient will be equal to 16.8, rounded 17 A, without the coefficient 21 A.
• We remember that we calculated the operating power parameters, but we also need to take into account the utmost valid value. To do this, we multiply the current strength we calculated by 1.4, because the correction for thermal effects is 40%. We got: 23.8 A and 29.4 A, respectively.
• So, in our example for safe work open wiring will require a cable with a cross-section of more than 3 mm², and for hidden option 2.5 mm².

Let's not forget that, due to various circumstances, we sometimes turn on more units at the same time than we expected. That there are also light bulbs and other devices that consume little energy. Let's stock up on some reserve section in case the fleet increases household appliances and with the calculations in hand, we’ll go make an important purchase.

Video guide for accurate calculations

Which cable is better to buy?

Following the strict recommendations of the PUE, for the arrangement of personal property we will buy cable products with the “letter groups” NYM and VVG in the marking. They are the ones who do not cause any complaints or quibbles from electricians and firefighters. Option NYM is an analogue of domestic VVG products.

It is best if the domestic cable is accompanied by the NG index, this means that the wiring will be fire resistant. If you plan to lay a line behind a partition, between joists or above suspended ceiling, buy products with low smoke emission. They will have the LS index.

This is a simple way to calculate the cross-section of the cable conductor. Information about the principles of calculations will help you rationally select this important element electrical networks. The necessary and sufficient size of the current-carrying core will provide power home appliances and will not cause a fire in the wiring.