The procedure for calculating and choosing the cross-section of power cables. WEBSOR Electrical Information Territory

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


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


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

Table for selecting cable cross-section for current and power with copper conductors
Copper conductors of wires and cables
Voltage, 220 V Voltage, 380 V
current, A power, kW current, A power, kW
1,5 19 4,1 16 10,5
2,5 27 5,9 25 16,5
4 38 8,3 30 19,8
6 46 10,1 40 26,4
10 70 15,4 50 33,0
16 85 18,7 75 49,5
25 115 25,3 90 59,4
35 135 29,7 115 75,9
50 175 38,5 145 95,7
70 215 47,3 180 118,8
95 260 57,2 220 145,2
120 300 66,0 260 171,6
Table for selecting cable cross-section for current and power with aluminum conductors
Cross-section of current-carrying conductor, mm 2 Aluminum conductors of wires and cables
Voltage, 220 V Voltage, 380 V
current, A power, kW current, A power, kW
2,5 20 4,4 19 12,5
4 28 6,1 23 15,1
6 36 7,9 30 19,8
10 50 11,0 39 25,7
16 60 13,2 55 36,3
25 85 18,7 70 46,2
35 100 22,0 85 56,1
50 135 29,7 110 72,6
70 165 36,3 140 92,4
95 200 44,0 170 112,2
120 230 50,6 200 132,0

Example of cable cross-section calculation

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

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

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

№ Number of veins
section mm.
Cables (wires) 		Outer diameter mm. Pipe diameter mm. Acceptable long
current (A) for wires and cables when laying: 		Permissible continuous current
for rectangular copper bars
sections (A) PUE
VVG VVGng KVVG KVVGE NYM PV1 PV3 PVC (HDPE) Met.tr. Du in the air in the ground Section, tires mm Number of buses per phase
1 1x0.75 2,7 16 20 15 15 1 2 3
2 1x1 2,8 16 20 17 17 15x3210
3 1x1.55,4 5,4 3 3,2 16 20 23 33 20x3275
4 1x2.55,4 5,7 3,5 3,6 16 20 30 44 25x3340
5 1x46 6 4 4 16 20 41 55 30x4475
6 1x66,5 6,5 5 5,5 16 20 50 70 40x4625
7 1x107,8 7,8 5,5 6,2 20 20 80 105 40x5700
8 1x169,9 9,9 7 8,2 20 20 100 135 50x5860
9 1x2511,5 11,5 9 10,5 32 32 140 175 50x6955
10 1x3512,6 12,6 10 11 32 32 170 210 60x61125 1740 2240
11 1x5014,4 14,4 12,5 13,2 32 32 215 265 80x61480 2110 2720
12 1x7016,4 16,4 14 14,8 40 40 270 320 100x61810 2470 3170
13 1x9518,8 18,7 16 17 40 40 325 385 60x81320 2160 2790
14 1x12020,4 20,4 50 50 385 445 80x81690 2620 3370
15 1x15021,1 21,1 50 50 440 505 100x82080 3060 3930
16 1x18524,7 24,7 50 50 510 570 120x82400 3400 4340
17 1x24027,4 27,4 63 65 605 60x101475 2560 3300
18 3x1.59,6 9,2 9 20 20 19 27 80x101900 3100 3990
19 3x2.510,5 10,2 10,2 20 20 25 38 100x102310 3610 4650
20 3x411,2 11,2 11,9 25 25 35 49 120x102650 4100 5200
21 3x611,8 11,8 13 25 25 42 60
rectangular copper bars
(A) Schneider Electric IP30
22 3x1014,6 14,6 25 25 55 90
23 3x1616,5 16,5 32 32 75 115
24 3x2520,5 20,5 32 32 95 150
25 3x3522,4 22,4 40 40 120 180 Section, tires mm Number of buses per phase
26 4x1 8 9,5 16 20 14 14 1 2 3
27 4x1.59,8 9,8 9,2 10,1 20 20 19 27 50x5650 1150
28 4x2.511,5 11,5 11,1 11,1 20 20 25 38 63x5750 1350 1750
29 4x5030 31,3 63 65 145 225 80x51000 1650 2150
30 4x7031,6 36,4 80 80 180 275 100x51200 1900 2550
31 4x9535,2 41,5 80 80 220 330 125x51350 2150 3200
32 4x12038,8 45,6 100 100 260 385 Permissible continuous current for
rectangular copper bars (A) Schneider Electric IP31
33 4x15042,2 51,1 100 100 305 435
34 4x18546,4 54,7 100 100 350 500
35 5x1 9,5 10,3 16 20 14 14
36 5x1.510 10 10 10,9 10,3 20 20 19 27 Section, tires mm Number of buses per phase
37 5x2.511 11 11,1 11,5 12 20 20 25 38 1 2 3
38 5x412,8 12,8 14,9 25 25 35 49 50x5600 1000
39 5x614,2 14,2 16,3 32 32 42 60 63x5700 1150 1600
40 5x1017,5 17,5 19,6 40 40 55 90 80x5900 1450 1900
41 5x1622 22 24,4 50 50 75 115 100x51050 1600 2200
42 5x2526,8 26,8 29,4 63 65 95 150 125x51200 1950 2800
43 5x3528,5 29,8 63 65 120 180
44 5x5032,6 35 80 80 145 225
45 5x9542,8 100 100 220 330
46 5x12047,7 100 100 260 385
47 5x15055,8 100 100 305 435
48 5x18561,9 100 100 350 500
49 7x1 10 11 16 20 14 14
50 7x1.5 11,3 11,8 20 20 19 27
51 7x2.5 11,9 12,4 20 20 25 38
52 10x1 12,9 13,6 25 25 14 14
53 10x1.5 14,1 14,5 32 32 19 27
54 10x2.5 15,6 17,1 32 32 25 38
55 14x1 14,1 14,6 32 32 14 14
56 14x1.5 15,2 15,7 32 32 19 27
57 14x2.5 16,9 18,7 40 40 25 38
58 19x1 15,2 16,9 40 40 14 14
59 19x1.5 16,9 18,5 40 40 19 27
60 19x2.5 19,2 20,5 50 50 25 38
61 27x1 18 19,9 50 50 14 14
62 27x1.5 19,3 21,5 50 50 19 27
63 27x2.5 21,7 24,3 50 50 25 38
64 37x1 19,7 21,9 50 50 14 14
65 37x1.5 21,5 24,1 50 50 19 27
66 37x2.5 24,7 28,5 63 65 25 38

It is necessary to select a cable cross-section for a voltage of 10 kV to power a 2TP-3 transformer substation with a power of 2x1000 kVA to power a slab warehouse at a metallurgical plant in the city of Vyksa, Nizhny Novgorod region. The power supply diagram is shown in Fig. 1. The length of the cable line from cell No. 12 is 800 m and from cell No. 24 is 650 m. The cables will be laid in the ground in pipes.

Table for calculating electrical loads according to 2TP-3

The three-phase short-circuit current in maximum mode on RU-10 kV buses is 8.8 kA. The protection duration, taking into account the complete shutdown of the circuit breaker, is 0.345 seconds. The cable line is connected to the switchgear through a VD4 type vacuum switch (Siemens).

The cross-section of a cable line for a voltage of 6 (10) kV is selected based on heating by the rated current, checked by thermal resistance to short-circuit currents, voltage losses in normal and post-emergency modes.

We choose a cable brand AABlu-10kV, 10 kV, three-core.

1. Determine the calculated current in normal mode (both transformers are turned on).

Where:
n – number of cables to the connection;

2. Determine the rated current in post-emergency mode, taking into account that one transformer is turned off:

3. We determine the economic section, according to the PUE section 1.3.25. The calculated current is taken for normal operation, i.e. the increase in current in post-emergency and repair modes of the network is not taken into account:

Jek =1.2 – the normalized value of the economic current density (A/mm2) is selected according to the PUE table 1.3.36, taking into account that the time of use of the maximum load Tmax = 6000 hours.

The cross section is rounded to the nearest standard 35 mm2.

Continuous permissible current for a cable with a cross-section of 3x35mm2 according to PUE, 7th ed. Table 1.3.16 is Id.t=115A > Icalc.av=64.9 A.

4. We determine the actually permissible current, in this case the condition Iph>Icalc.av. must be met:

The coefficient k1, which takes into account the temperature of the medium different from the calculated one, is selected according to table 2.9 [L1. from 55] and table 1.3.3 PUE. Considering that the cable will be laid in pipes in the ground. According to Table 2-9, the standard ambient temperature is +25 °C. The temperature of the cable cores is +65°C, in accordance with the PUE, ed. 7, clause 1.3.12.

Determined according to SNiP 01/23/99 table 3, actual temperature environment where the cable will be laid, in my case the city of Vyksa. Average annual temperature is - +3.8°C.

According to the PUE table 1.3.3, we select the coefficient k1 = 1.22.

Coefficient k2 – taking into account resistivity soil (taking into account geological surveys), selected according to the PUE 7th ed. table 1.3.23. In my case, the correction factor for normal soil with a resistivity of 120 K/W will be k2=1.

We determine the coefficient k3 according to the PUE table 1.3.26, taking into account the reduction in current load with the number of operating cables in one trench (in pipes or without pipes), taking into account that one cable is laid in one trench. We accept k3 = 1.

Having determined all the coefficients, we determine the actual permissible current:

5. We check the AABlu-10kV cable with a cross-section of 3x35mm2 for thermal stability in accordance with PUE clause 1.4.17.

  • Ik.z. = 8800 A - three-phase short-circuit current in maximum mode on RU-10 kV buses;
  • tl = tз + to.в =0.3 + 0.045 s = 0.345 s - protection duration taking into account complete shutdown of the circuit breaker;
  • tз = 0.3 s – longest protection time, in in this example the longest protection response time is in overcurrent protection;
  • tо.в = 45 ms or 0.045 s - total shutdown time of the VD4 type vacuum circuit breaker;
  • C = 95 - thermal coefficient under nominal conditions, determined from table. 2-8, for cables with aluminum conductors.

The cross section is rounded to the nearest standard 70 mm2.

6. Check the cable for voltage loss:

Where:
r and x - the values ​​of active and reactive resistances are determined according to table 2-5 [L1.s 48].

For a cable with aluminum conductors with a cross-section of 3x70mm2, active resistance r = 0.447 Ohm/km, reactance x = 0.086 Ohm/km.

We determine sinφ, knowing cosφ. Let's remember school course geometry.

If you do not know cosφ, you can determine for various electrical receivers by reference materials table 1.6-1.8 [L3, pp. 13-20].

6.2 In post-emergency mode:

From the calculations it is clear that the voltage losses in the line are insignificant, therefore, the voltage of consumers will practically not differ from the nominal one.

Thus, with the specified initial data, the AABlu-10 3x70 cable was selected.

To make cable selection easier, you can download all the literature that I used in this example in the archive.

Literature:

  • 1. Design cable networks and postings. Khromchenko G.E. 1980
  • 2. SNiP 23-01-99 Construction climatology. 2003
  • 3. Calculation and design of power supply systems for facilities and installations. Kabyshev A.V., Obukhov S.G. 2006
  • 4. Rules for the construction of electrical installations (PUE). Seventh edition. 2008

Often, before purchasing cable products, there is a need to independently measure its cross-section in order to avoid deception on the part of manufacturers, who, due to savings and setting a competitive price, may slightly underestimate this parameter.

It is also necessary to know how the cable cross-section is determined, for example, when adding a new energy-consuming point in rooms with old electrical wiring that does not have any technical information. Accordingly, the question of how to find out the cross-section of conductors always remains relevant.

General information about cable and wire

When working with conductors, it is necessary to understand their designation. There are wires and cables that differ from each other in their internal structure and technical characteristics. However, many people often confuse these concepts.

A wire is a conductor that has in its design one wire or a group of wires woven together and a thin common insulating layer. A cable is a core or a group of cores that has both its own insulation and a common insulating layer (sheath).

Each type of conductor will have its own methods for determining cross sections, which are almost similar.

Conductor materials

The amount of energy that a conductor transmits depends on a number of factors, the main one of which is the material of the current-carrying conductors. The following non-ferrous metals can be used as the core material of wires and cables:

  1. Aluminum. Cheap and lightweight conductors, which is their advantage. They are characterized by such negative qualities as low electrical conductivity, a tendency to mechanical damage, high transient electrical resistance of oxidized surfaces;
  2. Copper. The most popular conductors, which have a high cost compared to other options. However, they are characterized by low electrical and transition resistance at the contacts, fairly high elasticity and strength, and ease of soldering and welding;
  3. Aluminum copper. Cable products with aluminum cores coated with copper. They are characterized by slightly lower electrical conductivity than their copper counterparts. They are also characterized by lightness, average resistance and relative cheapness.

Important! Some methods for determining the cross-section of cables and wires will depend specifically on the material of their conductor component, which directly affects the throughput power and current strength (method of determining the cross-section of conductors by power and current).

Measuring the cross-section of conductors by diameter

There are several ways to determine the cross-section of a cable or wire. The difference in determining the cross-sectional area of ​​wires and cables will be that in cable products it is necessary to measure each core separately and summarize the indicators.

For information. When measuring the parameter under consideration with instrumentation, it is necessary to initially measure the diameters of the conductive elements, preferably removing the insulating layer.

Instruments and measurement process

The measuring instruments can be a caliper or a micrometer. Mechanical devices are usually used, but electronic analogues with a digital screen can also be used.

Basically, the diameter of wires and cables is measured using a caliper, since it is found in almost every household. It can also measure the diameter of wires in a working network, for example, a socket or panel device.

The diameter of the wire cross-section is determined using the following formula:

S = (3.14/4)*D2, where D is the diameter of the wire.

If the cable contains more than one core, then it is necessary to measure the diameter and calculate the cross-section using the above formula for each of them, then combine the result obtained using the formula:

Stotal= S1 + S2 +…+Sn, where:

  • Stotal – total area cross section;
  • S1, S2, …, Sn – cross sections of each core.

Just a note. To ensure the accuracy of the results obtained, it is recommended to take measurements at least three times, turning the conductor in different directions. The result will be the average.

In the absence of a caliper or micrometer, the diameter of the conductor can be determined using a regular ruler. To do this, you need to perform the following manipulations:

  1. Clean the insulating layer of the core;
  2. Twist the turns around the pencil tightly to each other (there should be at least 15-17 pieces);
  3. Measure the winding length;
  4. Divide the resulting value by the number of turns.

Important! If the turns are not laid evenly on the pencil with gaps, then the accuracy of the obtained results of measuring the cable cross-section by diameter will be in doubt. To increase the accuracy of measurements, it is recommended to take measurements with different sides. It will be difficult to wind thick wires onto a simple pencil, so it is better to resort to a caliper.

After measuring the diameter, the cross-sectional area of ​​the wire is calculated using the formula described above or determined using a special table, where each diameter corresponds to the cross-sectional area.

It is better to measure the diameter of a wire containing ultra-thin cores with a micrometer, since a caliper can easily break it.

The easiest way to determine the cable cross-section by diameter is using the table below.

Table of correspondence between wire diameter and wire cross-section

Diameter of conductor element, mmCross-sectional area of ​​the conductor element, mm2
0,8 0,5
0,9 0,63
1 0,75
1,1 0,95
1,2 1,13
1,3 1,33
1,4 1,53
1,5 1,77
1,6 2
1,8 2,54
2 3,14
2,2 3,8
2,3 4,15
2,5 4,91
2,6 5,31
2,8 6,15
3 7,06
3,2 7,99
3,4 9,02
3,6 10,11
4 12,48
4,5 15,79

Segment cable cross-section

Cable products with a cross-section of up to 10 mm2 are almost always produced in a round shape. Such conductors are quite sufficient to meet the domestic needs of houses and apartments. However, with a larger cross-section of the cable, the input cores from the external electrical network can be made in segment (sector) form, and it will be quite difficult to determine the cross-section of the wire by diameter.

In such cases, it is necessary to resort to a table where the size (height, width) of the cable takes the corresponding value of the cross-sectional area. Initially, it is necessary to measure the height and width of the required segment with a ruler, after which the required parameter can be calculated by correlating the obtained data.

Table for calculating the area of ​​an electric cable core sector

Cable typeSectional area of ​​the segment, mm2
S35 50 70 95 120 150 185 240
Four-core segmentV- 7 8,2 9,6 10,8 12 13,2 -
w- 10 12 14,1 16 18 18 -
Three-core segmental stranded, 6(10)V6 7 9 10 11 12 13,2 15,2
w10 12 14 16 18 20 22 25
Three-core segmental single-wire, 6(10)V5,5 6,4 7,6 9 10,1 11,3 12,5 14,4
w9,2 10,5 12,5 15 16,6 18,4 20,7 23,8

Dependence of current, power and core cross-section

It is not enough to measure and calculate the cross-sectional area of ​​the cable based on the diameter of the core. Before installing wiring or other types of electrical networks, you must also know throughput cable products.

When choosing a cable, you must be guided by several criteria:

  • the strength of the electric current that the cable will pass;
  • power consumed by energy sources;

Power

The most important parameter during electrical installation work (in particular, cable laying) is throughput. The maximum power of electricity transmitted through it depends on the cross-section of the conductor. Therefore, it is extremely important to know the total power of the energy consumption sources that will be connected to the wire.

Usually manufacturers household appliances, devices and other electrical products indicate on the label and in the documentation attached to them the maximum and average power consumption. For example, a washing machine can consume electricity ranging from tens of W/h during rinsing mode to 2.7 kW/h when heating water. Accordingly, a wire with a cross-section that is sufficient to transmit electricity of maximum power must be connected to it. If two or more consumers are connected to the cable, then the total power is determined by adding limit values each of them.

The average power of all electrical appliances and lighting devices in an apartment rarely exceeds 7500 W for a single-phase network. Accordingly, the cable cross-sections in the electrical wiring must be selected to this value.

So, for a total power of 7.5 kW, it is necessary to use a copper cable with a core cross-section of 4 mm2, which is capable of transmitting about 8.3 kW. The cross-section of the conductor with an aluminum core in this case must be at least 6 mm2, passing a current power of 7.9 kW.

In individual residential buildings, a three-phase power supply system of 380 V is often used. However, most equipment is not designed for such electrical voltage. A voltage of 220 V is created by connecting them to the network through a neutral cable with an even distribution of the current load across all phases.

Electric current

Often the power of electrical equipment and equipment may not be known to the owner due to the absence of this characteristic in the documentation or completely lost documents and labels. There is only one way out in such a situation - to calculate using the formula yourself.

Power is determined by the formula:

P = U*I, where:

  • P – power, measured in watts (W);
  • I – electric current strength, measured in amperes (A);
  • U is the applied electrical voltage, measured in volts (V).

When the strength of the electric current is unknown, it can be measured with instrumentation: an ammeter, a multimeter, a clamp meter.

After determining the power consumption and electric current, you can use the table below to find out the required cable cross-section.

Calculation of the cross-section of cable products based on current load must be carried out to further protect them from overheating. When too much electric current passes through conductors for their cross-section, destruction and melting of the insulating layer can occur.

The maximum permissible continuous current load is quantitative value electric current that can pass the cable for a long time without overheating. To determine this indicator, it is initially necessary to sum up the powers of all energy consumers. After this, calculate the load using the formulas:

  1. I = P∑*Ki/U (single-phase network),
  2. I = P∑*Kи/(√3*U) (three-phase network), where:
  • P∑ – total power of energy consumers;
  • Ki – coefficient equal to 0.75;
  • U – electrical voltage in the network.

Tablitz for matching the cross-sectional area of ​​copper conductorsconductor products current and power *

Section of cable and wire productsElectrical voltage 220 VElectrical voltage 380 V
Current strength, APower, kWCurrent strength, APower, kW
2,5 27 5,9 25 16,5
4 38 8,3 30 19,8
6 50 11 40 26,4
10 70 15,4 50 33
16 90 19,8 75 49,5
25 115 25,3 90 59,4
35 140 30,8 115 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

*Important! Conductors with aluminum conductors have different values.

Determination of the cable product in cross section - especially important process, in which miscalculations are unacceptable. You need to take into account all factors, parameters and rules, trusting only your calculations. The measurements taken must coincide with the tables described above - if they do not contain specific values, they can be found in the tables of many electrical engineering reference books.

Video

Current values ​​are easy to determine by knowing the rated power of consumers using the formula: I = P/220. Knowing the total current of all consumers and taking into account the ratio of the permissible current load for the wire (open wiring) per wire cross-section:

  • for copper wire 10 amperes per square millimeter,
  • for aluminum 8 amperes per square millimeter, you can determine whether the wire you have is suitable or whether you need to use another one.

When performing hidden power wiring (in a tube or in a wall), the given values ​​are reduced by multiplying by a correction factor of 0.8. It should be noted that open power wiring is usually carried out with a wire with a cross-section of at least 4 square meters. mm based on sufficient mechanical strength.

The above ratios are easy to remember and provide sufficient accuracy for using wires. If you need to know with greater accuracy the long-term permissible current load for copper wires and cables, then you can use the tables below.

The following table summarizes the data on power, current and cross-section of cable and conductor materials for calculations and selection of protective equipment, cable and conductor materials and electrical equipment.

Permissible long-term 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 unarmoured.

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

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


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

Summary table of wire cross-sections, current, power and load characteristics.

The table shows data based on the PUE for selecting cross-sections of cable and wire products, as well as rated and maximum possible currents of circuit breakers for single-phase household loads most often used in everyday life.


Smallest permissible cross-sections of cables and wires electrical networks in residential buildings.


  • Copper, U = 220 V, single phase, two-wire cable
  • Copper, U = 380 V, three phase, three-core cable

* cross-sectional value can be adjusted depending on the specific cable laying conditions


The smallest cross-sections of current-carrying conductors of wires and cables in electrical wiring.

Core cross-section, mm 2

Conductors

aluminum

Cords for connecting household electrical receivers

Cables for connecting portable and mobile power receivers in industrial installations

Twisted two-core wires with stranded cores for stationary installation on rollers

Unprotected insulated wires for fixed indoor electrical wiring:

directly on the bases, on rollers, clicks and cables

on trays, in boxes (except for blind ones):

single-wire

stranded (flexible)

on insulators

Unprotected insulated wires in external electrical wiring:

on walls, structures or supports on insulators;

overhead line inputs

under canopies on casters

Unprotected and protected insulated wires and cables in pipes, metal sleeves and blind boxes

Cables and protected insulated wires for stationary electrical wiring (without pipes, sleeves and blind boxes):

for conductors connected to screw terminals

for conductors connected by soldering:

single-wire

stranded (flexible)

Protected and unprotected wires and cables laid in closed channels or monolithic (in building structures or under plaster)

Today there is a wide range of cable products, with a cross-section of cores from 0.35 mm2. and higher.

If you choose the wrong cable cross-section for household wiring, the result can have two results:

  1. An overly thick core will “hit” your budget, because... its linear meter will cost more.
  2. If the conductor diameter is inappropriate (smaller than necessary), the conductors will begin to heat up and melt the insulation, which will soon lead to a short circuit.

As you understand, both the results are disappointing, so in front of and in the apartment it is necessary to correctly calculate the cable cross-section depending on the power, current strength and line length. Now we will look at each of the methods in detail.

Calculation of power of electrical appliances

For each cable there is a certain amount of current (power) that it can withstand when operating electrical appliances. If the current (power) consumed by all devices exceeds the permissible value for the conductor, then an accident will soon be unavoidable.

To independently calculate the power of electrical appliances in the house, you need to write down the characteristics of each appliance separately (stove, TV, lamps, vacuum cleaner, etc.) on a piece of paper. After this, all the values ​​are summed up and the resulting number is used to select a cable with cores with the optimal cross-sectional area.

The calculation formula looks like:

Ptotal = (P1+P2+P3+…+Pn)*0.8,

Where: P1..Pn – power of each device, kW

Please note that the resulting number must be multiplied by a correction factor of 0.8. This coefficient means that only 80% of all electrical appliances will work at the same time. This calculation is more logical, because, for example, you will definitely not use a vacuum cleaner or hair dryer for a long time without a break.

Tables for selecting cable cross-section by power:

These are reduced and simplified tables, more exact values you can find in paragraphs 1.3.10-1.3.11.

As you can see, for each specific type of cable the table values ​​have their own data. All you need is to find the nearest power value and look at the corresponding cross-section of the cores.

So that you can clearly understand how to correctly calculate the cable power, we will give a simple example:

We calculated that the total power of all electrical appliances in the apartment is 13 kW. This value must be multiplied by a factor of 0.8, which will result in 10.4 kW of actual load. Next in the table we look for suitable value in the column. We are satisfied with the figure “10.1” for a single-phase network (voltage 220V) and “10.5” if the network is three-phase.

This means that you need to choose a cross-section of cable cores that will power all the calculation devices - in an apartment, room or some other room. That is, such a calculation must be carried out for each outlet group powered from one cable, or for each device if it is powered directly from the panel. In the example above, we calculated the cross-sectional area of ​​the input cable cores for the entire house or apartment.

In total, we select a cross-section with a 6-mm conductor for a single-phase network or a 1.5-mm conductor for a three-phase network. As you can see, everything is quite simple and even a novice electrician can cope with this task on his own!

Current load calculation

Calculation of cable cross-section by current is more accurate, so it is best to use it. The essence is similar, but only in this case it is necessary to determine the current load on the electrical wiring. To begin with, we calculate the current strength for each of the devices using formulas.

If the house has a single-phase network, you must use the following formula for calculation:For a three-phase network, the formula will look like:Where, P – power of the electrical appliance, kW

cos Phi - power factor

More details about the formulas associated with calculating power can be found in the article:.

We draw your attention to the fact that the values ​​of the table values ​​will depend on the conditions for laying the conductor. At , the permissible current loads and power will be significantly greater than at .

Let us repeat, any cross-section calculation is carried out for a specific device or group of devices.

Table for selecting cable cross-section for current and power:

Calculation by length

Well, the last way to calculate the cable cross-section is by length. The essence of the following calculations is that each conductor has its own resistance, which contributes as the length of the line increases (the greater the distance, the greater the losses). In the event that the loss value exceeds 5%, it is necessary to choose a conductor with larger conductors.

The following methodology is used for calculations:

  • It is necessary to calculate the total power of electrical appliances and current strength (we provided the corresponding formulas above).
  • The electrical wiring resistance is calculated. The formula has next view: conductor resistivity (p) * length (in meters). The resulting value must be divided by the selected cable cross-section.

R=(p*L)/S, where p is the tabular value

We draw your attention to the fact that the length of current passage must be doubled, because The current initially flows through one core and then returns back through the other.

  • Voltage losses are calculated: the current is multiplied by the calculated resistance.

U losses =I load *R wires

LOSSES=(U losses /U nom)*100%

  • The amount of losses is determined: voltage losses are divided by the network voltage and multiplied by 100%.
  • The final number is analyzed. If the value is less than 5%, we leave the selected core cross-section. Otherwise, we select a “thicker” conductor.

Let's say we calculated that the resistance of our wires is 0.5 Ohm, and the current is 16 Amperes, then.



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