For a long and reliable cable service, it must be correctly selected and calculated. Electricians, when installing wiring, mostly choose the cross section of the wires, based mainly on experience. Sometimes this leads to errors. The calculation of the cable cross section is necessary, first of all, in terms of electrical safety. It will be wrong if the conductor diameter is smaller or larger than required.
Cable section too low
This case is the most dangerous, because the conductors overheat from the high current density, while the insulation melts and a short circuit occurs. This may also destroy electrical equipment, cause a fire, and workers may become energized. If you install a circuit breaker for the cable, it will work too often, which will create some discomfort.
Cable section is higher than required
Here the main factor is economic. The larger the cross section of the wire, the more expensive it is. If you do the wiring of the entire apartment with a large margin, it will cost a large amount. Sometimes it is advisable to make the main input of a larger cross section, if a further increase in the load on the home network is expected.
If you set the appropriate circuit breaker for the cable, the following lines will be overloaded when any of them does not trip their circuit breaker.
How to calculate cable size?
Before installation, it is advisable to calculate the cable cross-section according to the load. Each conductor has a certain power, which should not be less than that of the connected electrical appliances.
Power calculation
The easiest way is to calculate the total load on the input wire. The calculation of the cable cross-section according to the load is reduced to determining the total power of consumers. Each of them has its own denomination, indicated on the case or in the passport. Then the total power is multiplied by a factor of 0.75. This is due to the fact that all devices cannot be turned on at the same time. For the final determination of the required size, the cable section calculation table is used.
Calculation of the cable section by current
A more accurate method is the current load calculation. The cable cross-section is calculated by determining the current passing through it. For a single-phase network, the formula is applied:
Icalc.=P/(Unom∙cosφ),
where P - load power, Unom. - mains voltage (220 V).
If the total power of active loads in the house is 10kW, then the rated current Icalc.=10000/220 ≈ 46 A. When the cable cross-section is calculated by current, a correction is made for the conditions for laying the cord (indicated in some special tables), as well as to overload when turning on electrical appliances approximately upwards of 5 A. As a result, Icalc.=46 + 5=51 A.
The thickness of the cores is determined by the reference book. Calculation of the cable cross-section using tables makes it easy to find the right size for the continuous current. For a three-core cable laid into the house through the air, you must select a value in the direction of a larger standard section. It is 10mm2. The correctness of self-calculation can be checked by using an online calculator - cable section calculation, which can be found on some resources.
Cable heating during current flow
When the load is running, heat is generated in the cable:
Q=I2Rn w/cm, where I is the current, R is the electrical resistance, n is the number of cores.
From the expression it follows that the amount of power released is proportional to the square of the current flowing through the wire.
Calculation of the allowable current according to the heating temperature of the conductor
The cable cannot heat up indefinitely, as heat is dissipated into the environment. In the end, equilibrium occurs and a constant temperature of the conductors is established.
For a steady process, the ratio is true:
P=∆t/∑S=(tw - tav)/(∑S),
where ∆t=tw-tav - the difference between the temperature of the medium and the core, ∑S - temperature resistance.
The long-term permissible current passing through the cable is found from the expression:
Iadd=√((tadd - tav)/(Rn ∑S)),
where tadditional - allowable core heating temperature (depends on cable type and installation method). Usually it is 70 degrees in normal mode and 80 in emergency.
Conditions for heat dissipation with the cable running
When a cable is laid in an environment, heat dissipation is determined by its composition and humidity. The calculated resistivity of the soil is usually assumed to be 120 Ohm∙°C/W (clay with sand at a moisture content of 12-14%). To clarify, you should know the composition of the medium, after which you can find the resistance of the material according to the tables. To increase thermal conductivity, the trench is covered with clay. The presence of construction debris and stones in it is not allowed.
The heat transfer from the cable through the air is very low. It worsens even more when laying in a cable channel, where additional air layers appear. Here, the current load should be reduced compared to the calculated one. In the technical characteristics of cables and wires, the allowable short circuit temperature is given, which is 120 ° C for PVC insulation. Soil resistance is 70% of the total and is the main one in the calculations. Over time, the conductivity of the insulation increases as it dries out. This must be taken into account in the calculations.
Cable voltage drop
Due to the fact that the conductors have electrical resistance, part of the voltage is spent on heating them, and less comes to the consumer than it was at the beginning of the line. As a result, potential is lost along the length of the wire due to heat losses.
The cable must not only be selected according to the cross section to ensure its performance, but also take into account the distance over which energy is transmitted. An increase in load leads to an increase in current through the conductor. At the same time, losses increase.
Small voltage is applied to spotlights. If it decreases slightly, it is immediately noticeable. If you choose the wrong wires, the bulbs located farther from the power supply look 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 section of the cable is the consumer located farthest from the others. Losses are considered predominantly for this load.
On section L of the conductor, the voltage drop will be:
∆U=(Pr + Qx)L/Un,
where P and Q are active and reactive power, r and x are the active and reactance of section L, and Un- rated voltage at which the load normally operates.
Permissible ∆U from power sources to main inputs do not exceed ±5% for lighting residential buildings and power circuits. From the input to the load, the losses should not be more than 4%. For long lines, the inductive reactance of the cable must be taken into account, which depends on the distance between adjacent conductors.
Methods of connecting consumers
Loads can be connected in different ways. The most common are the following ways:
- at the end of the network;
- consumers are evenly distributed along the line;
- a line with uniformly distributed loads is connected to an extended section.
Example 1
The power of the appliance is 4 kW. Cable length is 20 m, resistivity ρ=0.0175 Ohm∙mm2.
The current is determined from the relationship: I=P/Unom=4∙1000/220=18.2 A.
Then, the cable section calculation table is taken and the appropriate size is selected. For a copper wire, it will be S=1.5 mm2.
Cable section calculation formula: S=2ρl/R. Through it, you can determine the electrical resistance of the cable: R=2∙0.0175∙20/1, 5=0.46 Ohm.
From the known value of R, we can determine ∆U=IR/U∙100 %=18.2100∙0.46/220∙100=3.8%.
The result of the calculation does not exceed 5%, which means that the losses will be acceptable. In case of large losses, it would be necessary to increase the cross-section of the cable cores by choosing the adjacent, larger size from the standard range - 2.5 mm2.
Example 2
Three lighting circuits are connected in parallel with each other on one phase of a three-phase load-balanced line, consisting of a four-wire cable 70 mm2 50 m long and carrying a current of 150 A. For eachlighting lines 20 m long carry a current of 20 A.
The phase-to-phase losses under the actual load are: ∆Uphase=150∙0.05∙0.55=4.1 V. Now you need to determine the loss between neutral and phase, since the lighting is connected to a voltage of 220 V: ∆Ufn=4, 1/√3=2, 36 V.
On one connected lighting circuit, the voltage drop will be: ∆U=18∙20∙0, 02=7, 2 V. Total losses are determined by the sum of Utotal=(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 during a long-term load, using tables, the cable cross-section is calculated according to the long-term permissible current. In addition, it is necessary to correctly calculate the wires and cables so that the voltage loss in them is not more than normal. At the same time, losses in the power circuit are summed up with them.
