Calculation of the cable section. Cable cross-section calculation table

For a long and reliable cable service, it must be properly selected and calculated. Electricians when installing wiring mostly choose the cross-section of cores, based mainly on experience. Sometimes this leads to errors. Calculation of the cable section is necessary, first of all, in terms of electrical safety. It will be wrong if the diameter of the conductor is less or more than required.

Cable cross-section is understated

This case is the most dangerous, since the conductors overheat from high current density, while the insulation melts and a short circuit occurs. In this case, electrical equipment may also collapse, a fire may occur, and workers may fall under tension. If you install a circuit breaker for the cable, it will operate too often, which will create some discomfort.

Cable cross-section higher than required

Here the main factor is economic. The larger the cross-section of the wire, the more expensive it is. If you make the posting of the whole apartment with a large margin, it will cost a large amount. Sometimes it is advisable to make the main input of a larger section if a further increase in the load on the home network is assumed.

If you install the appropriate machine for the cable, the following lines will be overloaded when one of them does not work its circuit breaker.

How to calculate the cross-section of a cable?

Before installation, it is advisable to calculate the cable cross-section by load. Each conductor has a certain power, which should not be less than that of the connected electrical appliances.

Power calculation

The simplest way is to calculate the total load on the lead-in wire. Calculation of the cable cross-section by load is reduced to determining the total power of consumers. Each of them has its own nominal value, indicated on the body or in the passport. Then, the total power is multiplied by a factor of 0.75. This is due to the fact that all instruments can not be switched on at the same time. For final determination of the required size, a cable cross-section calculation table is used.

Calculation of the current cable cross-section

A more accurate method is to calculate the current load. Calculation of the cable cross-section is made through the definition of the current flowing through it. For a single-phase network, the following formula applies:

I _calc. = P / (U _nom ∙ cosφ),

Where P is the load power, U _nom. - mains voltage (220 V).

If the total power of the active loads in the house is 10 kW, then the calculated current I _calc. = 10000/220 ≈ 46 A. When the current cable cross-section is calculated, a correction is made for the conditions for laying the cord (indicated in some special tables), as well as for overloading when turning on the electric appliances by about 5 A increments _. = 46 + 5 = 51 A.

The thickness of the veins is determined by reference. Calculation of the cable cross-section with the use of tables makes it easy to find the desired size for a long-time permissible current. For a three-core cable that is pushed through the house by air, you must select a value toward a larger standard section. It is 10 mm ² . The correctness of self-calculation can be checked by applying the online calculator - calculation of the cable cross-section, which can be found on some resources.

Cable heating with current passing

When the load is operating, the cable generates heat:

Q = I ² Rn W / cm,

Where I - current, R - electrical resistance, n - number of cores.

It follows from the expression that the amount of power output is proportional to the square of the current passing through the wire.

Calculation of the permissible current with respect to the temperature of the conductor heating

The cable can not be infinitely heated, since heat dissipates into the environment. Eventually, equilibrium sets in and a constant temperature of the conductors is established.

For the steady-state process, the following relationship holds:

P = Δt / ΣS = (t _ж - t _ср ) / (ΣS),

Where Δt = t _x -t _cp is the difference between the temperature of the medium and the core, and ΣS is the temperature resistance.

The long-term admissible current passing through the cable is found from the expression:

I _add = √ ((t _additional - t _cp ) / (RnΣS)),

Where t _additional - permissible temperature of core heating (depends on cable type and laying method). Usually it is 70 degrees in normal mode and 80 in emergency mode.

Heat dissipation conditions when the cable is running

When the cable is laid in any medium, the heat sink is determined by its composition and humidity. The calculated resistivity of the soil is usually taken to be 120 Ω ∙ ° C / W (clay with sand at a humidity of 12-14%). For specification it is necessary to know the composition of the medium, after which it is possible to find the resistance of the material from the tables. To increase the thermal conductivity, the trench is covered with clay. It is not allowed to have construction debris and stones in it.

The heat transfer from the cable through the air is very low. It deteriorates even more when laying in the cable channel, where additional air layers appear. Here, the current load should be reduced in comparison with the rated current. The technical characteristics of cables and wires lead to an acceptable short-circuit temperature of 120 ° C for PVC insulation. The soil resistance is 70% of the total and is the main one in the calculations. Over time, the conductivity of insulation increases due to its drying out. This must be taken into account in the calculations.

Voltage drop in the cable

Due to the fact that the conductors have electrical resistance, some of the voltage goes to their heating, and to the consumer it comes less than it was at the beginning of the line. As a result, the potential is lost along the length of the wire due to heat losses.

The cable should not only be selected by cross-section to ensure its operability, but also take into account the distance to which energy is transferred. Increasing the load leads to an increase in current through the conductor. In this case, the losses increase.

A small voltage is applied to the spotlights. If it decreases slightly, it is immediately noticeable. If the wires are incorrectly selected, the bulbs located further from the power supply unit appear dim. The voltage is significantly reduced in each subsequent section, and this is reflected in the brightness of the lighting. Therefore, it is necessary to calculate the cable cross-section along the length.

The most important part of the cable is the consumer located further than the rest. Losses are considered predominantly for this load.

In the section L of the conductor, the voltage drop is:

ΔU = (Pr + Qx) L / UH,

Where P and Q are the active and reactive power, r and x - the active and reactive resistance of the section L, and U _n - the nominal value of the voltage at which the load normally works.

The permissible ΔU from power supplies to main inputs does not exceed ± 5% for lighting of residential buildings and power circuits. From input to load, the loss should not be more than 4%. For lines with a long extension, it is necessary to take into account the inductive resistance of the cable, which depends on the distance between adjacent conductors.

Ways to connect consumers

Loads can be connected in different ways. The most common are the following methods:

• At the end of the network;
• Consumers are distributed along the line evenly;
• A line with uniformly distributed loads is connected to the extended section.

Example 1

The power of the appliance is 4 kW. The length of the cable is 20 m, the resistivity ρ = 0.0175 Ω ∙ mm ² .

The current is determined from the relationship: I = P / U _nom = 4 ∙ 1000/220 = 18.2 A.

Then the cable section calculation table is taken, and the appropriate size is selected. For copper wire, it will be S = 1.5 mm ² .

The formula for calculating the cable cross-section is: S = 2ρl / R. Through it, you can determine the electrical resistance of the cable: R = 2 ∙ 0.0175 ∙ 20 / 1.5 = 0.46 Ohm.

By the known value of R, it is possible to determine ΔU = IR / U ∙ 100% = 18.2 * 100 ∙ 0.46 / 220 ∙ 100 = 3.8%.

The result of the calculation does not exceed 5%, which means that the losses will be acceptable. In the case of large losses, it would be better to increase the cross-section of the cable cores by choosing a larger one from the standard series - 2.5 mm ² .

Example 2

The three lighting circuits are connected in parallel with each other to a single phase of a three-phase line balanced by loads consisting of a four-wire cable of 70 mm ² with a length of 50 m and conducting a current of 150 A. For each 20 m light line, a current of 20 A passes.

The phase-to-phase losses under the effective load are: ΔU _phases = 150 ∙ 0, 05 ∙ 0.55 = 4.1 V. Now it is necessary to determine the losses between the neutral and the phase, since the illumination is connected to 220 V: ΔU fn = 4 , 1 / √3 = 2.36 V.

On one connected lighting circuit, the voltage drop will be: ΔU = 18 ∙ 20 ∙ 0,02 = 7,2 V. The total losses are determined through the sum U _total = (2,4 + 7,2) / 230 ∙ 100 = 4,2 %. The calculated value is below the allowable loss, which is 6%.

Conclusion

To protect the conductors from overheating with a long-term load using the tables, the cable cross-section is calculated for the long-term admissible current. In addition, you need to correctly calculate the wires and cables, so that the voltage loss in them is no more than the norm. In this case, the losses in the power circuit are summed up with them.