The cut-off cross-section of the cable is standardized in all countries. This applies to both the CIS countries and Europe. This issue is regulated in our country by the document “Rules for the Construction of Electrical Installations,” which are called PUE. Calculation of cable cross-section by power is selected using special tables. Of course, many calculate the necessary conductor parameters “by eye,” but this is not entirely correct. This indicator may be different for each apartment. This is due to the number of electrical consumers and their power. Without proper calculation, many unpleasant situations can occur, costly repairs of both wiring and apartments.
Cable device
To determine the power cross-section of a cable, you should understand its principle and design. It can be compared, for example, with a water or gas pipeline. In the same way as through these communications, a flow flows through an electrical conductor. Its power limits the cross-section of the conductor.
The cable cross-section using the power indicator can be performed incorrectly in two cases:
- The current carrying channel will be too narrow. This will lead to an increase in current density and, consequently, to overheating of the insulation. This state of the conductor over time will be characterized by the presence weak points where leakage is possible. This condition of the channel can cause a fire.
- The current-carrying wire is too wide. This is certainly not the worst option. The spaciousness of transportation of the electrical flow will allow for more functional and durable use of the conductor. However, as the cross-section increases, the cost of the cable also increases.
The first option poses a danger to life, health and property. The second method is safe, but the materials will be quite expensive to purchase.
Easy way
Calculation of cable cross-section by power is based on the well-known law developed by Ohm. It tells you that the current flow multiplied by the voltage will equal the power. Voltage in everyday life is considered a constant value. In a single-phase network it is equal to 220 V. Therefore, to determine the cable cross-section based on current and power, only two variables remain.
Next, the current values and the expected load are calculated. Moreover, the cable size based on its power can be selected according to the PUE table. This indicator is calculated for the wire suitable for the sockets. Traditionally, for lighting lines, a wire with a cross section of 1.5 mm 2 is laid.
However, it happens that devices such as a hair dryer, microwave, electric kettle, etc. are connected to the socket group. It is necessary to distribute the load and correctly calculate the cable cross-section based on power indicators, correlating the diameter and the load.
If it is not possible to separate the socket groups, then many electricians will recommend immediately installing a cable with a copper core of up to 6 mm 2.
Sectional area and diameter
Calculation of cable cross-section by power, diameter and load are not equivalent concepts. The first indicator is calculated in mm 2, and the second - simply in mm. You can select the power and permissible current according to the table both by the cross-section of the cable and by its diameter.
If the table takes into account only the size of the cross-sectional area in mm 2, and there is data only on the cable diameter, the missing indicator can be found using the following formula:
S = 3.14D2/4 = 0.785D2,
where: S is the wire cross-section, and D is the diameter.
If the cross-section of the wire is not round, but rectangular, its cross-sectional area is calculated by multiplying the length by the width (just like the area of a rectangle).
Load Based Calculation
The easiest way to calculate the cable cross-section is by summing the powers of all units that will be connected to the line. To do this, you will need to perform a certain sequence of actions.
First, it is determined which electrical appliances will be used in the home, which of them will likely function simultaneously. Next, you need to look at the technical data sheets of each of these units. It will be necessary to calculate the sum of the powers of those electrical consumers that must operate simultaneously.
Then the figure obtained as a result of the calculations is rounded up. This will ensure a safe supply of power for the electrical wiring. The cross-section of the wire or cable is calculated further using PUE tables.
In a similar way, you can summarize the current strength, which is indicated in the data sheets of electrical equipment. Rounding and searching is performed using the power calculation table.
Table of power, current and cross-section of copper wires
According to the PUE, in residential buildings it is necessary to use only copper conductors for wiring. The power supply for some electrical equipment, which belongs to engineering types of receivers, can be connected to the network using aluminum conductors with a cross-section of at least 2.5 mm 2.
Table of power, current and cross-section of aluminum wires
The specialist will also be able to make correction factors based on the type of wire location, temperature environment, for a cable in the ground, etc. The table for calculating cable power, cross-section or current applies to conductors in plastic or rubber insulation. These include such common brands as GDP, PVS, PPV, VPP, AVVG, VVG, APPV, etc. Non-insulated or paper-screened cables must be calculated according to the table relating to them.
Length and section
Calculation of the cable cross-section by power simply must be used to determine its length. This data is important when creating long extension cords. The exact values obtained must be increased by 10-15 cm. This margin is necessary for switching using soldering, welding or crimping.
In construction, the cable cross-section is calculated based on power and length at the electrical wiring design stage. This is very important, especially for those communications that will be subject to significant or additional loads.
In everyday life, the wire length is calculated using the following formula:
I=P/U*cosφ, where:
- P - power (W);
- I - current strength (A);
- U - voltage (V);
- cosφ is a coefficient that is equal to 1.
The cable cross-section must first be found in the table. The formula will help determine correct length wires.
Current Density
The current strength varies in the range of 6-10 A, which was determined experimentally. This value is calculated for the current flowing through 1 mm 2 of copper conductor.
This statement means that the calculation of the cable cross-section for power and current takes as a basis a copper cable with a cross-sectional area of 1 mm 2, through which a current of 6 to 10 A can flow without melting or overheating to waiting household electrical appliances.
According to the PUE code, for each wire a reserve of 40% is allocated for overheating that is safe for the sheath. If a value of 6 A characterizes the operation of the presented conductor for an infinitely long term without time limits, then a value of 10 A is suitable for the short-term flow of current through the core.
If a current of 12 A flows through a 1 mm 2 copper conductor, it will be cramped in such a conductor. This will lead to an increase in current density. The core will begin to heat up and melt the insulation.
Therefore, such calculations are required when choosing a cable cross-section for each type of wiring.
Having familiarized yourself with the methods that allow you to calculate the cable cross-section based on power and current, you can install or repair old wiring that will last a long time and will be completely safe for people living in the house. Many quite simple, but effective ways will help you determine exactly required size sections for the electrical network.
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 power cable- at the entrance. 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 on all cases household appliances and in their passports it is indicated.
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 section larger size. Due to this, a reserve is created for the subsequent commissioning of additional electrical appliances, which may be acquired 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 installation.
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. Usually it is intended for the kitchen, where it is connected electric stove. 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 the entire twist, 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
More exact result gives a calculation of the cable cross-section based on 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 fluorescent lamps and motors of electrical appliances (refrigerator, vacuum cleaner, power tools, etc.).
Current example
Let's find out what to do if you need to determine the cross-section of the copper cable to connect household appliances with a total power of 25 kW and three-phase machines 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 %.
Carrying Suitable for welding machine according to 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.
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 must not be laid 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 strength 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. This will be a 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. Are used in alternating circuits with a rated voltage of 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.
The PVS installation wire is intended for connecting 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 AC installations. 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 domestic stationary installation indoors and 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))
In this article I will tell you how choose the right cable section for a house or apartment. If- this is the “heart” of our power supply system, then the cables connected to the electrical panel circuit breakers are"blood vessels" that supplyelectricity from our household electrical receivers.
When installing electrical wiring in a house or apartment, all stages, starting from designing the electrical supply of a private house or apartment, and ending with the final installation of sockets or switches, must be approached with full responsibility, because your personal electrical safety, as well as the fire safety of your house or apartment, depends on it . Therefore, we approach the choice of cable cross-section with all seriousness, because no other method of transmitting electricity in a private house or apartment has yet been invented.
It is important to choose the correct cable cross-section specifically for a specific line (group) of electrical receivers. Otherwise, if we choose a lower section cable is will lead to overheating, destruction of insulation and further to a fire If you touch a cable with damaged insulation, you will receive an electric shock. If you choose a cable cross-section that is too high for a house or apartment, this will lead to increased costs, and difficulties will arise during the electrical installation of cable lines, because the larger the cable cross-section, the more difficult it is to work with it; not every socket will “fit” a cable with a cross-section of 4 sq. mm .
I bring common universal table, which I myself use to select the rated current of circuit breakers for cable protection.blue
I won’t fill your head with abstruse formulas for calculating cable cross-section from books on electrical engineering, so that you can choose the right cable cross-section. Everything has long been calculated and tabulated.
Please note that when in different ways installation of electrical wiring(hidden or open) , cables with the same cross-section have different continuous-permissible currents.
Those. at open way installation of electrical wiring, the cable heats up less due to better cooling. At h covered way installation of electrical wiring (in grooves, pipes, etc.), on the contrary, it heats up more. This is an important point, because if you choose the wrong machine to protect the cable, the rating of the machine may turn out to be overestimated relative to the long-term permissible current of the cable, which is why the cable can get very hot, but the machine will not turn off.
I'll bring you example, for example, we have a cable cross-section of 6 sq. mm:
- at open method its continuous-permissible current is 50A, therefore the machine must be set to 40A;
- with the hidden method, its long-term permissible current is 34A, in this case the machine is 32A.
Suppose we chose a cable cross-section for an apartment, which are laid in grooves or under plaster (in a closed way). If we mix it up and install 50A circuit breakers for protection, the cable will overheat, because at closed method its gasket In = 34 A, which will lead to destruction of its insulation, then a short circuit and fire.
TABLES ARE NOT CURRENT. WHEN SELECTING A MACHINE FOR CABLES, USE THE TABLE ABOVE.
Cable cross-section for hidden electrical wiring
Cable cross-section for open electrical wiring
To use the tables and choose the right cable cross-section for a house or apartment, we need to know the current strength, or know the power of all household electrical receivers.
The current is calculated using the following formulas:
— for a single-phase network with a voltage of 220 Volts:
where P is the sum of all powers of household electrical receivers, W;
U - single-phase network voltage 220 V;
Cos(phi) - power factor, for residential buildings it is 1, for production it will be 0.8, and on average 0.9.
— for a three-phase network with a voltage of 380 Volts:
in this formula everything is the same as for a single-phase network, only in the denominator, because The network is three-phase, add root 3 and the voltage will be 380 V.
To select the cable cross-section for a house or apartment, according to the above tables, it is enough to know the sum of the powers of electrical receivers for a given cable line(groups). We will still need current calculations when designing an electrical panel (choosing automatic machines, RCDs or differential circuit breakers).
Below are the average power values of the most common household electrical appliances:
Knowing the power of electrical receivers, you can accurately select the cable cross-section for a specific cable line (group) in a house or apartment, and therefore an automatic device (difavtomatic) to protect this line, whose rated current must be lower than the continuous-permissible current of the cable of a certain cross-section. If we choose a copper cable cross-section of 2.5 sq. mm, which conducts current up to 21 A for as long as desired ( hidden installation method), then the automatic circuit breaker (difavtomat) in the electrical panel for this cable must have a rated current of 20 A so that the circuit breaker turns off before the cable begins to overheat.
Typical cable sections for household electrical installations:
- In apartments, cottages or private houses, for socket groups laying copper cable 2.5 sq.mm.;
- For lighting group- copper cable cross-section 1.5 sq.mm;
- For single phase hob (electric stoves) - cable section 3x6 sq. mm., for a three-phase electric stove - 5x2.5 sq. mm. or 5x4 sq. mm. depending on power;
- For other groups (ovens, boilers, etc.) - by their power. And also on the connection method, through a socket or through terminals. For example, if the power oven more than 3.5 kW, then lay a 3x4 cable and connect the oven through the terminals; if the oven power is less than 3.5 kW, then a 3x2.5 cable and connection through a household outlet is sufficient.
To select the correct cable cross-section and ratings of circuit breakers for the electrical panel of a private house, apartment, you need to know important points , not knowing which can lead to dire consequences.
For example:
- For socket groups, select a cable cross-section of 2.5 sq. mm, but the machine is chosen with a rated current of not 20A, but 16A, because household sockets designed for a current of no more than 16 A.
- For lighting I use a 1.5 sq. mm cable, but automatic machine no more than 10A, because switches are designed for a current of no more than 10A.
- You need to know that the machine passes current up to 1.13 times its nominal value, for as long as you like, and if the nominal value is exceeded up to 1.45 times, it can turn off only after 1 hour. And all this time the cable will heat up.
- Select the correct cable cross-section according to in a hidden way gaskets so that there is the necessary margin of safety.
- PUE clause 7.1.34. prohibits the use aluminum wiring inside buildings.
Thank you for your attention.
When repairing and designing electrical equipment, it becomes necessary to choose correctly. You can use a special calculator or reference book. But for this you need to know the load parameters and cable laying features.
TO electrical networks the following requirements apply:
- safety;
- reliability;
- efficiency.
If the selected cross-sectional area of the wire is small, then the current loads will not be large, which will lead to overheating. As a result, there may be emergency, which will harm 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 right choice wire cross-section is the key to long-term safe operation and rational use of financial resources.
A separate chapter in the PUE is devoted to the correct selection of a conductor: “Chapter 1.3. Selection of conductors based on heating, economic current density and corona conditions.”
The cable cross-section is calculated based on power and current. Let's look at examples. To determine what wire gauge is needed for 5 kW, you will need to use PUE tables (“ Rules for electrical installations“). 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 ( kW).
- Cable location.
There is no meaning in the PUE 5 kW, so you have to choose the next larger value -- 5.5 kW. For installation in an apartment today it is necessary. 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 it is designed. introductory machine (VA). According to " Rules for electrical installations“, 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 VA. In this case, after 25 A there is 35 A. The last value must be taken as the calculated value. A current of 35 A corresponds to a cross section of 4 mm² and a power of 7.7 kW. So, the choice of section copper wire by power completed: 4 mm².
To find out what wire gauge is needed for 10 kW, let's use the reference book again. 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 will be 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 of the devices; 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, 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 |
| 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, 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 | 132,2 |
In addition, you need to know the network voltage: three-phase corresponds to 380 V, and single-phase corresponds to 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.
Lack of copper conductors-- high cost. In Soviet houses, aluminum electrical wiring was used during construction. Therefore if it happens partial replacement, then it is advisable to install aluminum wires. The only exceptions are those cases when, instead of all the old wiring ( to switchboard ) a new one is installed. 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 provided for room temperature (18°C). To select the cable cross-section by current, use the PUE tables (PUE-7 clause 1.3.10-1.3.11 PERMISSIBLE CONTINUOUS CURRENTS FOR WIRES, CORDS AND CABLES WITH RUBBER OR PLASTIC INSULATION).
Table 3. Electric current for copper wires and cords with rubber and PVC insulation
| Conductor cross-sectional area, mm² | ||||||
| open | in one pipe | |||||
| 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.
Table 4. 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 | |||||
| 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 | - | - | - | - | - |
Except 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. 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 whose dimensions are 1 square millimeter by 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 power P, voltage U and coefficient cosф we find the current using the formula: I=P/(U*cosф). For electrical networks used in everyday life, cosф = 1. In industry, cosph 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 conductor resistance 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 is: R = Ro*2.
- We find the voltage drop from the relationship: ΔU=I*R.
- Determine the voltage drop as a percentage: ΔU/U. If the obtained value exceeds 5%, then select the nearest larger one from the directory cross section conductor.
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.