Table of loads on the cable section: selection, calculation

The comfort and safety in the home depend on the correct choice of the cross-section of the wiring. When overloaded, the conductor overheats, and the insulation may melt, resulting in a fire or short circuit. But the cross section is more than necessary to take unprofitable, because 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 applies a table of loads along the cable section. Using 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.

cable cross-section load table

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

cable calculation

With an overestimation of the cross section of the wire, it will cost more. Although a certain margin is necessary, since in the future, as a rule, it is necessary to connect new equipment. It is advisable to apply a safety factor of the order of 1.5.

Total power calculation

The total power consumed by the apartment falls on the main input, which enters the switchboard, and after it branches out on the line:

  • lighting;
  • outlet groups;
  • separate powerful electrical appliances.

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

power cable section

All capacities add up. Similarly, calculations are made for each circuit. Experts propose to multiply the amount by a reduction factor of 0.75. This is due to the fact that at the same time all devices are not connected to the network. Others suggest choosing a larger section. Due to this, a reserve is created for the subsequent commissioning of additional electrical appliances that may be acquired in the future. It should be noted that this option of cable calculation is more reliable.

cable diameter

How to determine the cross section of the wire?

In all calculations, the cable cross section appears. Its diameter is easier to determine if you apply the formula:

  • S = π D² / 4 ;
  • D = √ (4 × S / π).

Where π = 3.14.

cable core section

In a stranded wire, you first need to count the number of wires (N). Then, the diameter (D) of one of them is measured, after which the cross-sectional area is determined:

S = N × D² / 1.27.

Stranded wires are used where flexibility is required. Cheaper one-piece conductors are used for fixed installation.

How to choose a cable by power?

In order to select the wiring, the table of loads on the cable section is used:

  • If the open type line is under voltage 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 wiring lighting.
  • With a power of 6 kW, larger conductors are required - 2.5 mm². The wire is used for outlets 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 an electric stove is connected. A similar load is supplied via a separate line.

Which cables are better?

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

cable brands

The wire is available in monolithic and multiwire. Each core, as well as the entire strand, is insulated from the outside with PVC, and the filler between them is non-combustible:

  • So, the NUM cable is used indoors, since the insulation on the street is destroyed by sunlight.
  • And as an internal and external wiring widely used cable brand VVG. It is cheap and reliable enough. For laying in the ground it is not recommended to be used.
  • Wire brand VVG is made flat and round. No filler is used between the cores.
  • The VVGNG-P-LS cable is made with an external sheath that does not support combustion. The cores are made round to a cross section of 16 mm², and over - sector.
  • The brands of PVA and ShVVP cables are made multi-wire and are used mainly for connecting household appliances. It is often used as a home wiring. On the street, multi-wire cores are not recommended because of corrosion. In addition, bending insulation cracks at low temperatures.
  • Armored and moisture resistant cables AVBShv and VBShv are laid underground under the street. The armor is made of two steel strips, which increases the reliability of the cable and makes it resistant to mechanical stress.

Current Load Detection

A more accurate result is given by calculating the cable cross-section by power and current, where the geometric parameters are connected with the electric ones.

cable cross sections for power and current

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

I = P / (U ∙ cosφ).

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

Example of calculating the cable cross section for current

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

Given the reactive component, the power of household appliances and equipment will be:

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

Input currents are determined:

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

If we distribute single-phase loads evenly across three phases, the current will fall on one:

I f = 162/3 = 54 A.

At each phase there will be a current load:

I f = 54 + 38 = 92 A.

All equipment at the same time will not work. Given the margin for each phase, the current is:

I f = 92 × 0.75 × 1.5 = 103.5 A.

In a five-core cable, only phase conductors are taken into account. For a cable laid in the ground, it is possible to determine for a current of 103.5 A the cross-section of conductors 16 mm (table of loads on the cable section).

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

Cable voltage drop

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

  1. The resistance of the conductor is determined: R = 2 × (ρ × L) / S.
  2. There is a voltage drop: U pad. = I × R. In relation to the linear percentage, it will be: U % = (U pad. / U lin. ) × 100.

The following notation is used in the formulas:

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

Coefficient 2 shows that the current flows through two cores.

Example of calculating the cable according to the voltage drop

For example, it is necessary to calculate the voltage drop across the carrier with a core section of 2.5 mm², 20 m long. It is necessary for connecting a welding transformer with a power of 7 kW.

  • 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.
  • Carrying voltage drop: U pad. = 31.8 × 0.28 = 8.9 V.
  • Percentage of voltage drop: U % = (8.9 / 220) × 100 = 4.1 %.

Carrying is suitable for the welding machine according to the requirements of the rules of operation of electrical installations, since the percentage of voltage drop on it is within normal limits. However, its value on the supply wire remains large, which may adversely affect the welding process. Here it is necessary to check the lower permissible supply voltage limit for the welding machine.

Conclusion

In order to reliably protect the wiring from overheating during prolonged exceeding the rated current, the cable cross-sections are calculated according to the long-term permissible currents. The calculation is simplified if a table of loads on the cable section is used. A more accurate result is obtained if the calculation is performed at the maximum current load. And for stable and long-term operation, a circuit breaker is installed in the wiring circuit.


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