Summary of the article:
What factors affect the cross section?
Choose the right cable section for electrical wiring affect factors such as:
- conductive material;
- insulation material;
- laying method;
- total power of connected devices.
Similar material is set out in 9 minute video:
For electrical wiring in residential buildings recommend use wires and cables with copper cores.
7th edition of the “Rules for the Installation of Electrical Installations” clause 7.1.34 on page 411 forbids the use of aluminum conductors (there are exceptions).
Why? Aluminum compared to copper:
- conducts current 1.64 times worse (has a specific resistance of 0.02828 Ohm · mm 2 / m compared with 0.01724 Ohm · mm 2 / m);
- when closed, it can “leak” from the contact joint due to the low melting point (657 and 1083 ° C, respectively);
- fragile (with several bends, the core will break);
- causes overheating of contact compounds due to the formation of an oxide film, which has a high transition resistance.
The operating current of the wiring is limited by the maximum allowable temperature of the heating of the core when current flows through it.
In wiring more often use PVC insulation (the optimal ratio of price and maximum heating temperature), its maximum heating temperature with a rated current of 70 ° C. When this temperature is exceeded for a certain time, varying degrees of change occur:
|Core and insulation temperature, ° С||Exposure time||Consequences for isolation|
|80||> 8 ocloc’k||deformed|
|160||> 4 seconds||cracking|
- GOST 16442-80 standard of tables 8 and 9 on page 7, clauses 2.6.4 and 2.6.5 on page 9, tables 21 and 22 on page 16;
- GOST standard 53769-2010 table 18 on page 24;
- GOST standard 7399-97 table 2 on page 3 (last column).
Conductors are placed in cable channels, corrugated pipe, rigid pipe or strobe. All these laying methods in PUE-7 are united by a single term. in a pipe.
Total power of connected devices
power requires a larger cross-section of the core. Examples of power of electrical appliances and reciprocal sections below.
If we compare the current power with the volume of water in the pipe, then the more water you need to “drive”, the larger the diameter of the pipe is needed.
Comprehensive selection of core and circuit breaker
Consider electrical wiring in apartments, houses, cottages, cottages, garages and other domestic and household buildings (single-phase network with a modular input circuit breaker for current ≤100A without powerful electric motors). To create it, a two-core or three-core copper cable is used (phezanol or phase-earthing, respectively).
Table selection of the optimal section veins proceeding for practical reasons:
|Appointment||Cross section of one core, mm 2|
|powerful electrical appliances (stove, washing machine, water heater and so on)||4-10 (depending on the power of the device)|
And now justify it.
|Permissible continuous current for wires and cords with rubber and PVC insulation with copper conductors|
veins, mm 2
|Current in amperes for wires laid:|
|open||in one pipe|
|two single-core||three single-core||four single-core||one two-core||one three-core|
|Cross section of one core, mm 2||Permissible current Icab, AND|
- Iav. rated current of the circuit breaker, which is necessary to protect the wiring;
- Icab. permissible current per cable core;
- 1.45. coefficient selected based on table 5.
|Time-current characteristics thermal release circuit breaker|
|Test current||Tripping||Off time
|1.13 · In||no||t > 60 min at In 120 min at In > 80A|
|1.45 · In||there is||t 80A|
|2.55 · In||there is||1 s 40A|
Video: What Cable Section For Trimmer is 1 Kilowatt
Partial reproduction of table 6 on page 19 of the standard for modular circuit breakers GOST 50345.
For instance, if to protect a wire with a cross-section of 4 mm 2 we install a 32 amp circuit breaker (according to table 4 we see that the conductor can withstand the same continuous current), then when overloaded, the circuit breaker will not immediately turn off and will pass 32 · 1 current for an hour, 45 = 46.4 A. During this time, the insulation of the wire will get overheated, which is fraught with cable failure.
To ensure safety (margin of “strength”), reliability and durability of electrical wiring for everyday use, the optimum ratio of the cross section of the used wire and the rating of the circuit breaker to protect the laid cable is as in table 6.
The calculated current is determined by the formula (1), rated current of the machine choose the closest to the calculated from the existing line of rated currents: 3, 4, 5, 6, 8, 10, 13, 16, 20, 25, 32, 40, 50, 63, 80, 100 A.
|Cross section of one core, mm 2||Permissible current Icab, AND||The rated current of the machine, A||Rated current of the machine Iav, AND|
or Iav = 13 amperes, if you find it on sale (then the calculated power will be higher by 30%). In the IEK nomenclature there are such and we supply.
can I takeav = 25 amperes, then the calculated power will be higher by 25%.
Next, we determine the power according to the formula:
- P is the total power of the connected devices;
- Iav. rated current of the machine from table 6;
- U. alternating voltage in the network (220 volts),
and write the results in table 7.
|Cross section of one core, mm 2||Machine current Iav, AND||Ultimate power P, W||Rounded value of P, kW|
- 1.5 mm 2. 1 and 2 kW;
- 2.5 mm 2. 3 kW;
- 4 mm 2. 4 kW;
- 6 mm 2. 5 kW;
- 10 mm 2. 6, 7 and 8 kW;
- 16 mm 2. 9, 10 and 11 kW.
For lighting we use a wire with a cross section of 1.5 mm 2, protected by an automatic machine of 10 A. A load of 2.2 kW is allowed on such a wire, that is, it will be enough for 22 incandescent lamps of 100 W each, which is more than enough for an apartment.
For indoor outlets, which are planned to include appliances low power (phone, tablet, laptop, computer, speakers. all these consumers are up to 100 W) we use one a wire with a cross section of 2.5 mm 2, protected by an automatic machine of 16 A. The principle is the same, the total power of the devices included in the socket, and therefore should not exceed 3.5 kW, applied to the cable.
Important rule: to 1 outlet (with 1-5 places for plugs) we connect 1 cable protected by 1 circuit breaker.
If you want to connect two or more outlets to one cable with a cross section of 2.5 mm2, then with the inclusion of an iron with a power of 2.1 kW and a room heater with a power of 2.0 kW, the total load on the cable will be 4.1 kW, which is clearly more than permissible 3.5 kW (circuit breaker trips).
In the case of one double outlet, when you want to turn on the iron and heater together, you will think about it. is it worth it?
For direct connection of powerful appliances or for outlets in which they will obviously be connected (for example, an outlet in the bathroom to connect a washing machine or boiler), use cables with a section of 4, 6 or 10 mm 2, and it is mandatory separate cable to an individual consumer. We select the cross section according to the same principle. The power of a particular device is indicated in its factory passport, however, you can navigate according to table 8 (it is also part of Appendix 7.2 in RM-2696-01).
- Current loads in DC networks
- Electrical resistivity of some metals used in electrical engineering
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Cable section. This is the cutting area of the current-carrying core. If the cut of the vein is round (as in most cases) and consists of one wire. then the area / section is determined by the circle area formula. If there are a lot of wires in a vein, then the section will be the sum of the sections of all the wires in this vein.
The cross-sectional values in all countries are standardized, and the standards of the former CIS and Europe in this part are completely the same. In our country, the document that regulates this issue are "Electrical Installation Rules" or briefly. PUE.
The cable section is selected based on the loads using special tables called "Permissible current loads on the cable." If there is no desire to understand these tables. then it’s quite enough for you to know that it is advisable to take copper cables with a cross section of 1.5-2.5 mm² for sockets, and for lighting. 1.0-1.5mm². 6.0mm² is enough to enter one phase into an ordinary 2-3 room apartment. Anyway, your 40-80 m² larger equipment does not fit, even taking into account the electric stove.
Many electricians for "outfits" They believe that 1 mm² of copper wire can pass 10A of electric current through itself: respectively, 2.5 mm² of copper can pass 25A, and 4.0 mm². 40A, etc. If you analyze the table for selecting the cable section a little, you will see that this method is suitable only for estimation and only for cables with a section of no higher than 6.0 mm².
The following is an abbreviated table for selecting a cable cross-section up to 35 mm² depending on current loads. There, for your convenience, the total power of electrical equipment is shown for 1-phase (220V) and 3-phase (380V) consumption. Please note that when laying the cable in the pipe (i.e. in any enclosed spaces, such as in the wall), the possible current loads on the cable should be less than when laying open. This is due to the fact that the cable heats up during operation, and the heat transfer in the wall or in the ground is much lower than in open space.
Important When the load is called in kW. this is the total load. Those. for a single-phase consumer, the load will be indicated in one phase, and for a three-phase consumer. collectively for all three. When the magnitude of the load is named in amperes (A). it’s always about the load on one core (or phase).
|Cable cross-section, mm²||Paved open||Laid in the pipe|
|current, A||kW||current, A||kW||current, A||kW||current, A||kW|
If you carefully studied the above table and still want to independently determine the cable section you need, for example, for entering the house, then you also need to know the following. This table applies to cables and wires in rubber and plastic insulation. These are such widespread brands as: PVA, GDP, runway, PPV, APPV, VVG. AVVG and a number of others. The cable in paper insulation has its own table, on non-insulated wires and busbars. its. When calculating the cable cross-section, the specialist must also take into account the cable laying methods: in trays, bundles, etc. In addition, the values from the tables on permissible current loads should be adjusted by the following reducing factors:
correction factor corresponding to the cable cross-section and its location in the block;
correction factor for ambient temperature;
correction factor for cables laid in the ground;
correction factor for a different number of working cables laid side by side.
If this does not stop you. then open the directory under the editorship of Belorussov on page 503, and we take off our hat.
If money is not a problem for you, then feel free to increase the reference cross section of the vein by 50%, and sleep peacefully: since even all the correction factors in the amount will not give more.
When calculating the required cable section, the main criterion. this is the amount of heat generated by the cable when an electric current passes through it and the ambient temperature. In fact, any electrical conductor can pass a lot of current through itself, up to its melting point, and this is ten times more than indicated in the directories. Please note that the reference values are for long current loads on the cable. And short-term loads can be much higher. Those. stock is always there. But provided that you purchased a cable manufactured in accordance with GOST. If instead of a copper cable you were sold something made of some alloy and coated with plastic from recycled polyethylene (from used bags and PET bottles), then why do you need all these tables: see article "How to choose a cable"
Current loads in DC networks
In DC networks, the cross section calculation is somewhat different. The resistance of the conductor to DC voltage is much higher than AC (with alternating current, resistance at lengths up to 100 m is generally neglected). In addition, for DC consumers, as a rule, it is very important that the voltage at the ends is not lower than 0.5V (for AC consumers, as you know, voltage fluctuations within the range of 10% are acceptable in any order). There is a formula that determines how much the voltage at the ends drops compared to the base voltage, depending on the length of the conductor, its specific resistance and current strength in the circuit:
U. DC voltage, V
p. wire resistivity, Ohm 2 / m
l. wire length m
S. cross-sectional area, mm 2
I. current strength, A
Knowing the values of these indicators, it is quite easy to calculate the section you need: by the substitution method, or by using simple arithmetic operations on this equation.
If the drop in the DC voltage at the ends does not matter, then to select the cross section, you can use the table for alternating current, but at the same time adjust the current values by 15% in the direction of decrease, i.e. at constant current, cable reference sections can pass current by 15% less than indicated in the table. A similar rule also works for the selection of circuit breakers for DC networks, for example: for circuits with a load of 25A, you need to take the machine 15% lower than the rated value, in our case the previous machine size is suitable. 20A.
Electrical resistivity of some metals used in electrical engineering
|Metal||Resistance, Ohm · mm 2 / m|
Attention: This is an author’s article, therefore, when using the material, please make a reference to the source.