Maximum allowable current for copper wires

When an electric current flows through the cable, some of the energy is lost. It goes to the heating of the conductors because of their resistance, with a decrease in which the value of transmitted power and the permissible current for copper wires increase. The most acceptable conductor in practice is copper, which has a small electrical resistance, suits consumers at a cost and is available in a wide range.

The next metal with good conductivity is aluminum. It is cheaper than copper, but more brittle and deformed at joints. Previously, domestic domestic networks were laid with aluminum wires. They were hidden under plaster and for a long time forgot about electroconducting. Electricity mostly went to the lighting, and the wires easily withstand the load.

With the development of technology, many electrical appliances appeared, which became indispensable in everyday life and required more electricity. The power consumption increased and the wiring ceased to cope with it. Now it became unthinkable to do electricity supply to an apartment or a house without calculating the wiring for capacity. Wires and cables are chosen so that there is no extra cost, and they are fully able to cope with all the loads in the house.

The reason for heating the wiring

The passing electrical current causes the conductor to heat up. At an elevated temperature, the metal rapidly oxidizes, and the insulation begins to melt at a temperature of 65 ^° C. The more often it heats up, the faster it breaks down. For this reason, the wires are selected according to the admissible current, at which they do not overheat.

Section area of the wiring

The shape of the wire is a circle, a square, a rectangle or a triangle. In apartment wiring, the section is predominantly circular. A copper bus is usually installed in a distribution cabinet and is rectangular or square.

The cross-sectional areas of the veins are determined by the main dimensions measured by a caliper:

• Circle - S = πd 2/4;
• The square is S = a ² ;
• Rectangle - S = a * b;
• The triangle is πr 2/3.

The following designations are used in the calculations:

• R is the radius;
• D is the diameter;
• B, a is the width and length of the section;
• Π = 3.14.

Calculation of power in the wiring

The power released in the veins of the cable during its operation is determined by the formula: P = I _n ² Rn,

Where I _n - load current, A; R - resistance, Ohm; N is the number of conductors.

The formula is suitable for calculating one load. If several cables are connected to the cable, the amount of heat is calculated separately for each energy consumer, and then the results are summed up.

The permissible current for copper stranded wires is also calculated through the cross section. To do this, you need to fluff the end, measure the diameter of one of the wires, calculate the area and multiply by their number in the wire.

Wire cross-section for different operating conditions

Wire cross-sections are conveniently measured in square millimeters. If we roughly estimate the permissible current, mm2 of the copper wire passes through itself 10 A, without overheating.

In the cable, the adjacent wires heat each other, so for him it is necessary to choose the core thickness according to the tables or with the correction. In addition, the sizes are taken with a small margin in the direction of increase, and then selected from the standard series.

The wiring can be open and hidden. In the first variant, it is laid outside on surfaces, in pipes or in cable channels. Hidden passes under the plaster, in channels or pipes inside the structures. Here the operating conditions are more stringent, since in closed spaces without air access the cable heats up more strongly.

For different operating conditions, correction factors are introduced to which the calculated long-time permissible current should be multiplied depending on the following factors:

• Single-core cable in a pipe longer than 10 m: I = I _n x 0.94;
• Three single-core cables in one pipe: I = I _n x 0.9;
• A gasket in water with a protective coating of the type Кл: I = I _n х 1.3;
• Four-core cable of equal cross-section: I = I _n x 0.93.

Example

With a load of 5 kW and a voltage of 220 V, the current through the copper wire will be 5 x 1000/220 = 22.7 A. Its cross section will be 22.7 / 10 = 2.27 mm ² . This size will provide an acceptable current for copper wires by heating. Therefore, here you should take a small margin of 15%. As a result, the cross section will be S = 2.27 + 2.27 x 15/100 = 2.61 mm ² . Now, to this size, you should choose the standard cross-section of the wire, which is 3 mm.

Heat dissipation during cable operation

The conductor can not be heated from the passing current indefinitely. At the same time, it gives heat to the environment, the amount of which depends on the temperature difference between them. At a certain moment, an equilibrium state sets in and the conductor temperature is set constant.

Important! With properly selected wiring, the heat loss is reduced. It should be remembered that for irrational consumption of electricity (when the wires overheat) also have to pay. On the one hand, the fee is charged for the extra expense of the meter, and on the other hand, for the replacement of the cable.

Wire section selection

For a typical apartment, the electricians do not particularly think about which sections of the wiring to choose. In most cases, the following are used:

• Lead-in cable - 4-6 mm ² ;
• Sockets - 2,5 mm ² ;
• The main lighting is 1.5 mm ² .

Such a system copes well with the loads, if there are no powerful electrical appliances, which sometimes need to maintain a separate power supply.

Great for finding the permissible current of a copper wire, the table from the directory. It also shows the calculation data for the use of aluminum.

The basis for choosing the wiring is the power of consumers. If the total power in the lines from the main input P = 7.4 kW at U = 220 V, the permissible current for copper wires will be according to Table 34A, and the cross-section - 6 mm ² (closed gasket).

Short-time operation modes

The maximum permissible short-time current for copper wires with operating modes with a cycle duration of up to 10 minutes and operating periods between them of not more than 4 minutes is reduced to a long operating mode if the cross-section does not exceed 6 mm ² . With a cross section above 6 mm ² : I _add = I _n ∙ 0.875 / √T _ae. ,

Where Т _п.в - the ratio of the duration of the working period to the duration of the cycle.

The power failure during overloads and short circuits is determined by the technical characteristics of the used protective devices. Below is a diagram of a small control panel of the apartment. The meter feeds to the MCA DP MCB with a power rating of 63 A, which protects the wiring before the machines of individual lines with a capacity of 10 A, 16 A and 20 A.

Important! The thresholds for the operation of automatic machines must be less than the maximum permissible current of the wiring and above the load current. In this case, each line will be reliably protected.

How to choose the right input wire to the apartment?

The value of the rated current on the cable entering the apartment depends on how many consumers are connected. The table shows the required instruments and their power.

Electrical Appliance	Rated power, kW
TV	0.18
Boiler	2-6
Fridge	0.2-0.3
Oven	2-5
A vacuum cleaner	0.65-1
Electric kettle	1.2-2
Iron	1.7-2.3
Microwave oven	0.8-2
A computer	0.3-1
Washer	2.5-3.5
Lighting system	0.5
Total	12.03-23.78

The current strength from a known power can be found from the expression:

I = P ∙ K _and / (U ∙ cos φ), where K _and = 0.75 is the simultaneity coefficient.

For most electrical appliances that are an active load, the power factor is cos φ = 1. For luminescent lamps, electric vacuum cleaner motors, washing machines, etc., it is less than 1 and must be taken into account.

The long-lasting current for the devices shown in the table will be I = 41 - 81 A. The value is quite impressive. You should always think carefully when you buy a new electrical appliance, whether its apartment network will pull. According to the table for open wiring the cross-section of the input wire will be 4-10 mm ² . Here still it is necessary to consider, how the apartment load will affect общеобмовую. It is possible that the Housing Office will not allow the connection of as many electrical appliances to the riser, where a bus (copper or aluminum) passes through the switchboards for each phase and neutral. They simply do not pull the electricity meter, which is usually installed in the shield on the landing. In addition, the charge for the overrun of the electricity norm will grow to impressive proportions because of the rising coefficients.

If the wiring is done for a private house, then here it is necessary to take into account the power of the outgoing wire from the main network. The commonly used aluminum wire SIP-4 with a cross section of 12 mm ² may not be enough for a large load.

The choice of wiring for individual consumer groups

Once the cable is selected for connection to the network and the input device protecting it from overloads and short-circuits is selected, it is necessary to select wires for each group of consumers.

Load is divided into lighting and power. The most powerful consumer in the house is the kitchen, where an electric stove, a washing machine and a dishwasher, a refrigerator, a microwave oven and other electrical appliances are installed.

For each outlet, 2.5 mm ² wires are selected. According to the table for concealed wiring, he will miss 21 A. The supply scheme is usually radial - from the junction box. Therefore, the wires must be 4 mm ² wide. If the sockets are connected by a loop, it should be taken into account that the cross section of 2.5 mm ² corresponds to a power of 4.6 kW. Therefore, the total load on them should not exceed it. There is one drawback: if one outlet fails, the rest can also be inoperative.

On a boiler, electric stove, air conditioning and other powerful loads, it is advisable to connect a separate wire with an automatic device. In the bathroom there is also a separate input with automatic and RCD.

The lighting is led to a wire of 1.5 mm ² . Now many people use basic and additional lighting, where a larger cross section may be required.

How to calculate three-phase wiring?

The type of network influences the calculation of the permissible cable cross-section . If the power consumption is the same, the permissible current loads on the core of the cable for a three-phase network will be less than for single-phase.

To supply a three-core cable at U = 380 V, the following formula is used:

I = P / (√3 ∙ U ∙ cos φ).

The power factor can be found in the characteristics of electrical appliances or it is equal to 1, if the load is active. The maximum permissible current for copper wires, as well as aluminum for three-phase voltages, is indicated in the tables.

Conclusion

To prevent overheating of the conductors under prolonged load, the cross-section of the conductors must be correctly calculated, on which the permissible current for the copper wires depends. If the power of the conductor is not enough, the cable will prematurely fail.