Load table by cable section: choice, calculation

From the correct choice of the section of the electrical wiring depends the comfort and safety in the house. When the load is overloaded, the conductor overheats, and the insulation may melt, resulting in a fire or short circuit. But the cross section is more disadvantageous, since the price of the cable increases.

In general, it is calculated depending on the number of consumers, for which first determine the total power used by the apartment, and then multiply the result by 0.75. In the PUE, a table of loads along the cable section is used. It can easily determine the diameter of the wires, which depends on the material and the passing current. As a rule, copper conductors are used.

The cross-section of the cable core should exactly match the calculated conductor - in the direction of increasing the standard size range. The most dangerous when it is understated. Then the conductor constantly overheats, and the insulation quickly fails. And if you install an appropriate circuit breaker, then it will often occur.

If the cross section of the wire is too high, it will cost more. Although a certain stock is necessary, as in the future, as a rule, it is necessary to connect new equipment. It is advisable to apply 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 out on the line:

• lighting;
• Groups of sockets;
• Separate powerful electrical appliances.

Therefore, the largest section of the power cable is at the input. On the outgoing 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 passports to them.

All capacities are added up. Similarly, calculations are made for each contour. Experts propose to multiply the amount by a decreasing factor of 0.75. This is due to the fact that all devices are not included in the network at the same time. Others suggest choosing a larger section. This creates a reserve for the subsequent commissioning of additional electrical appliances that can be purchased in the future. It should be noted that this version of cable calculation is more reliable.

How to determine the cross-section of a wire?

In all calculations, the cross-section of the cable appears. By its diameter, it is easier to determine if the following formulas are used:

• S = π D² / 4 ;
• D = √ (4 × S / π).

Where π = 3.14.

In a stranded wire, you first need to calculate 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. The cheaper one-piece conductors are used for stationary installation.

How to choose a cable by power?

In order to select the wiring, the table of loads by the cable section is applied:

• If the line of open type is under voltage 220 V, and the total power is 4 kW, a copper conductor of 1.5 mm² cross section is taken. This size is usually used for lighting wiring.
• At a power of 6 kW, veins with a larger cross section of 2.5 mm² are required. The wire is used for outlets to which household appliances are connected.
• The power of 10 kW requires the use of a wiring of 6 mm². Usually it is intended for the kitchen, where an electric stove is connected. Approach to such a load is made on a separate line.

Which cables are better?

Electricians know the German NUM cable for office and residential premises. In Russia, cable brands are produced that are lower in performance, although they may have the same name. They can be distinguished by the inflow of the compound in the space between the veins or by its absence.

The wire is manufactured in monolithic and multiwire. Each vein, as well as the entire twist on the outside, is insulated with PVC, the filler between them being incombustible:

• So, the NUM cable is used indoors, because the insulation on the street is destroyed by the sun's rays.
• And as an internal and external electrical wiring is widely used cable brand BBG. It is cheap and reliable enough. For laying in the ground it is not recommended to use.
• Wire VVG brand is made flat and round. Between the veins, the filler is not applied.
• The VVGng-P-LS cable is made with an outer jacket that does not support combustion. The veins are made round to a cross-section of 16 mm², and over - the sector.
• The brands of PVS and SHVVP cables are made of multiwire and are used mainly for connecting household appliances. It is often used as a domestic electrical wiring. In the street, stranded cores should not be used because of corrosion. In addition, the bending insulation crack at low temperature.
• On the street under the ground lay armored and moisture-resistant cables АВБШв and ВБШв. Armor is made of two steel bands, which increases the reliability of the cable and makes it resistant to mechanical stress.

Determination of the current load

A more accurate result is a calculation of the cable cross-section for power and current, where the geometric parameters are related to electrical.

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

I = P / (U ∙ cosφ).

Reactive load creates luminescent lamps and engines of electrical appliances (refrigerator, vacuum cleaner, power tool, etc.).

Example of calculating the cross-section of a current cable

Let's find out what to do if you need to determine the cross-section of a copper cable for connection of household appliances with a total power of 25 kW and three-phase machines by 10 kW. This connection is made by a five-core cable laid in the ground. The food of the house is made from a three-phase network.

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 _обор. = 10 / 0.7 = 14.3 kW.

The currents at input are defined:

• I _life. = 35.7 × 1000/220 = 162 A;
• I _obor. = 14.3 × 1000/380 = 38 A.

If we distribute single-phase loads evenly in three phases, one will have a current:

I _φ = 162/3 = 54 A.

Each phase will have a current load:

I _ф = 54 + 38 = 92 A.

All equipment at the same time will not work. Taking into account the margin for each phase, there is a current:

I _φ = 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, it is possible to determine the cross-section of conductors of 16 mm² for a current of 103.5 A (Table of loads along the cable section).

A refined calculation of the current strength makes it possible to save costs, since a smaller cross section is required. With a rougher cable calculation for power, the cross-section of the core will be 25 mm2, which will cost more.

Voltage drop on the cable

Conductors have resistance, which must be taken into account. This is especially important for long cable lengths or for small cross sections. PES norms are established, according to which the voltage drop on the cable should not exceed 5%. The calculation is done as follows.

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

The following formulas are used in the formulas:

• Ρ - specific resistance, Ohm × mm² / m;
• S is the cross-sectional area, mm².

The coefficient 2 shows that the current flows through two cores.

Example of cable calculation for voltage drop

For example, it is necessary to calculate the voltage drop on the carry with the cross section of the conductor 2.5 mm², length 20 m. It is necessary to connect a welding transformer with a power of 7 kW.

• The resistance of the wire is: R = 2 (0.0175 × 20) / 2.5 = 0.28 Ω .
• Current strength in the conductor: I = 7000/220 = 31.8 A.
• Voltage drop on the carry: 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 operating rules of electrical installations, since the percentage of voltage drop on it is within the normal range. However, its value on the supply wire remains large, which can adversely affect the welding process. Here it is necessary to check the lower permissible voltage limit for the welding machine.

Conclusion

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