Maximum allowable current for copper wires

When an electric current flows through the cable, part of the energy is lost. It is spent on heating the conductors because of their resistance, with a decrease in which the value of the 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 deforms at joints. Previously, domestic domestic networks were laid with aluminum wires. They were hidden under the plaster and for a long time forgot about the wiring. Electricity predominantly went into lighting, and the wires easily withstood the load.

With the development of technology, many electrical appliances have appeared that have become indispensable in everyday life and require more electricity. Power consumption increased and wiring ceased to cope with it. Now it has become unthinkable to make power supply to an apartment or house without calculating the wiring for power. Wires and cables are selected so that there is no unnecessary costs, and they completely cope with all the loads in the house.

Reason for heating wiring

The passing electric current causes the conductor to heat up. At elevated temperatures, the metal quickly oxidizes, and the insulation begins to melt at a temperature of 65 ⁰ C. The more often it heats up, the faster it breaks down. For this reason, the wires are selected according to the permissible current at which they do not overheat.

Cross-sectional Area

In shape, the wire is made in the form of a circle, square, rectangle or triangle. In apartment wiring, the cross section is predominantly round. The copper busbar is usually installed in the control cabinet and can be rectangular or square.

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

• circle - S = πd 2/4;
• square - S = a ² ;
• rectangle - S = a * b;
• the triangle is πr 2/3.

The following notation is 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 cores of the cable during its operation is determined by the formula: P = I _n ² Rn,

where I _n is the load current, A; R is the resistance, Ohm; n is the number of conductors.

The formula is suitable for calculating a single load. If several of them are connected to the cable, the amount of heat is calculated separately for each energy consumer, and then the results are summarized.

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 various operating conditions

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

In the cable, adjacent wires warm each other, so for it you need to choose the thickness of the core according to the tables or as amended. In addition, the dimensions are taken with a small margin in the direction of increase, and then choose from the standard series.

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

For different operating conditions, correction factors are introduced by which the calculated long-term allowable 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;
• gasket in water with a protective coating of type C: I = I _n x 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 acceptable current for copper heating wires. Therefore, a small margin of 15% should be taken here. As a result, the cross section is S = 2.27 + 2.27 x 15/100 = 2.61 mm ² . Now, to this size, you should choose a standard section of the wire, which will be 3 mm.

Heat dissipation during cable operation

The conductor cannot heat up from the passing current for an infinitely long time. At the same time, it gives off 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 temperature of the conductor is fixed.

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

Wire cross section selection

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

• input cable - 4-6 mm ² ;
• sockets - 2.5 mm ² ;
• main lighting - 1.5 mm ² .

A similar system can cope with the loads if there are no powerful electrical appliances, to which it is sometimes necessary to conduct a separate power supply.

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

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

Short-term operating modes

The maximum permissible short-term current for copper wires under operating conditions with a cycle duration of up to 10 minutes and working periods between them of no more than 4 minutes is reduced to a continuous mode of operation if the cross section does not exceed 6 mm ² . When the cross section is above 6 mm ² : I _add = I _n ∙ 0.875 / √ T a.p. ,

where T _p.v - the ratio of the duration of the working period to the duration of the cycle.

Power off during overloads and short circuits is determined by the technical characteristics of the used circuit breakers. Below is a diagram of a small control panel of an apartment. Power from the meter is supplied to the 63 MC DP MCB input circuit breaker, which protects the wiring of individual lines with 10 A, 16 A and 20 A power to the machines.

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

How to choose the lead wire to the apartment?

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

The current strength by 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 active loads, the power factor cos φ = 1. For fluorescent lamps, electric motors of a vacuum cleaner, a washing machine, etc. it is less than 1 and must be taken into account.

The long-term allowable 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 purchasing a new appliance whether the apartment network will pull it. According to the table for open wiring, the cross section of the input wire will be 4-10 mm ² . Here it is also necessary to take into account how the housing load will affect the common house. It is possible that the housing office will not allow connecting so many electrical appliances to the riser of the staircase, where a busbar (copper or aluminum) passes through the distribution cabinets for each phase and neutral. They simply cannot be pulled by an electric meter, which is usually installed in a switchboard on the landing. In addition, the charge for overspending the norm of electricity will grow to an impressive size due to increasing factors.

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

Posting selection for individual consumer groups

After you select a cable to connect to the network and select an input circuit breaker that protects against overloads and short circuits, you must select the wires for each group of consumers.

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

For each outlet, wires of 2.5 mm ² are selected. According to the table for hidden wiring, he will skip 21 A. The supply scheme is usually radial - from the junction box. Therefore, 4 mm ² wires should fit the box. If the sockets are connected by a loop, it should be borne in mind that a 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 others may also be inoperative.

It is advisable to connect a separate wire with an automatic machine to the boiler, electric stove, air conditioner and other powerful loads. A separate entry with an automatic machine and an RCD is also made into the bathroom.

A 1.5 mm ² wire is used for lighting. Now many use primary and secondary lighting, where a larger cross section may be required.

How to calculate three-phase wiring?

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

To supply a three-core cable at U = 380 V, the 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 at three-phase voltage, is indicated in the tables.

Conclusion

To prevent overheating of the conductors during prolonged load, it is necessary to correctly calculate the cross-section of the conductors, on which the permissible current for copper wires depends. If the power of the conductor is insufficient, the cable will prematurely fail.