Calculation of cable cross-section by power and current: how to correctly calculate wiring

Are you planning to do modernization of the power grid or additionally extend the power line to the kitchen to connect a new electric stove? Here, minimal knowledge about the cross-section of the conductor and the effect of this parameter on power and current will be useful.

Agree that incorrect calculation of the cable cross-section leads to overheating and short circuit or to unjustified costs.

It is very important to carry out calculations at the design stage, since failure hidden wiring and subsequent replacement is associated with significant costs. We will help you understand the intricacies of calculations in order to avoid problems in the further operation of electrical networks.

In order not to burden you with complex calculations, we have selected clear formulas and calculation options, presented the information in an accessible form, and provided the formulas with explanations. Thematic photos and video materials have also been added to the article, allowing you to clearly understand the essence of the issue under consideration.

Calculation of cross-section for consumer power

The main purpose of conductors is to deliver electrical energy to consumers in the required quantity. Since superconductors are not available under normal operating conditions, the resistance of the conductor material must be taken into account.

Calculation of the required section conductors and cables depending on the total power of consumers is based on long-term operating experience.

Let's begin the general course of calculations by first carrying out calculations using the formula:

P = (P1+P2+..PN)*K*J,

Where:

• P – the power of all consumers connected to the calculated branch in Watts.
• P1, P2, PN – power of the first, second, nth consumer, respectively, in Watts.

Having received the result at the end of the calculations using the above formula, it was time to turn to the tabular data.

Now you have to select the required section according to Table 1.

Table 1. The wire cross-section must always be selected to the nearest larger side (+)

Stage #1 - calculation of reactive and active power

Consumer capacities are indicated in the equipment documents. Typically, equipment data sheets indicate active power along with reactive power.

Devices with an active type of load convert all received electrical energy, taking into account efficiency, into useful work: mechanical, thermal or another type.

Devices with active loads include incandescent lamps, heaters, and electric stoves.

For such devices, the calculation of power by current and voltage has the form:

P=U*I,

Where:

• P – power in W;
• U – voltage in V;
• I – current strength in A.

Devices with a reactive type of load are able to accumulate energy coming from a source and then return it. This exchange occurs due to the displacement of the current sinusoid and the voltage sinusoid.

At zero phase displacement, power P=U*I always has a positive value. Devices with an active type of load have such a graph of the phases of current and voltage (I, i - current, U, u - voltage, π - pi equal to 3.14)

Devices with reactive power include electric motors, electronic devices of all sizes and purposes, and transformers.

When there is a phase shift between the current sinusoid and the voltage sinusoid, the power P=U*I can be negative (I, i - current, U, u - voltage, π - pi equal to 3.14). A reactive power device returns stored energy back to the source

Electrical networks are built in such a way that they can transfer electrical energy in one direction from source to load.

Therefore, the returned energy from a consumer with a reactive load is parasitic and is wasted on heating the conductors and other components.

Reactive power depends on the phase angle between the voltage and current sinusoids. The phase shift angle is expressed through cosφ.

To find the total power, use the formula:

P = Q / cosφ,

Where Q – reactive power in VAR.

Typically, the device data sheet indicates reactive power and cosφ.

Example: the passport for the hammer drill indicates a reactive power of 1200 VAr and cosφ = 0.7.Therefore, the total power consumption will be equal to:

P = 1200/0.7 = 1714 W

If cosφ could not be found, for the vast majority of household electrical appliances cosφ can be taken equal to 0.7.

Stage #2 - search for simultaneity and margin coefficients

K – dimensionless simultaneity coefficient, shows how many consumers can be connected to the network at the same time. It rarely happens that all devices consume electricity at the same time.

Simultaneous operation of the TV and music center is unlikely. From established practice, K can be taken equal to 0.8. If you plan to use all consumers simultaneously, K should be set to 1.

J – dimensionless safety factor. Characterizes the creation of a power reserve for future consumers.

Progress does not stand still; every year new amazing and useful electrical devices are invented. Electricity consumption is expected to grow by 84% by 2050. Typically J is taken to be between 1.5 and 2.0.

Stage #3 - performing calculations using the geometric method

In all electrical calculations, the cross-sectional area of ​​the conductor is taken - the cross-section of the core. Measured in mm².

It is often necessary to learn how to correctly calculate wire diameter conductor wires.

In this case, there is a simple geometric formula for a monolithic round wire:

S = π*R² = π*D²/4, or vice versa

D = √(4*S / π)

For rectangular conductors:

S = h * m,

Where:

• S – core area in mm²;
• R – core radius in mm;
• D – core diameter in mm;
• h, m – width and height, respectively, in mm;
• π — pi equals 3.14.

If you purchase a stranded wire, in which one conductor consists of many twisted wires of round cross-section, then the calculation is carried out according to the formula:

S = N*D²/1,27,

Where N – number of wires in the core.

Wires with cores twisted from several wires generally have better conductivity than monolithic ones. This is due to the peculiarities of current flow through a conductor with a round cross-section.

Electric current is the movement of like charges along a conductor. Like charges repel each other, so the charge distribution density is shifted towards the surface of the conductor.

Another advantage of stranded wires is their flexibility and mechanical resistance. Monolithic wires are cheaper and are mainly used for stationary installation.

Stage #4—calculate the power cross-section in practice

Task: the total power of consumers in the kitchen is 5000 W (meaning that the power of all reactive consumers has been recalculated). All consumers are connected to a single-phase 220 V network and are powered from one branch.

Table 2. If you plan to connect additional consumers in the future, the table shows the required power of common household appliances (+)

Solution:

Let us take the simultaneity coefficient K equal to 0.8. The kitchen is a place of constant innovation, you never know, the safety factor is J=2.0. The total estimated power will be:

P = 5000*0.8*2 = 8000 W = 8 kW

Using the value of the calculated power, we look for the nearest value in Table 1.

The closest suitable core cross-section for a single-phase network is a copper conductor with a cross-section of 4 mm². Similar wire size with 6mm aluminum core².

For single-core wiring, the minimum diameter will be 2.3 mm and 2.8 mm, respectively.In the case of using a multi-core option, the cross-section of the individual cores is summed up.

Calculation of current cross-section

Calculations of the required current and power cross-section of cables and wires will provide more accurate results.Such calculations make it possible to evaluate the overall influence of various factors on conductors, including thermal load, brand of wires, type of laying, operating conditions, etc.

The entire calculation is carried out in the following steps:

• selection of power of all consumers;
• calculation of currents passing through a conductor;
• selection of a suitable cross-section using tables.

For this calculation option, the power of consumers in terms of current and voltage is taken without taking into account correction factors. They will be taken into account when summing the current strength.

Stage #1 - calculation of current strength using formulas

For those who have forgotten the school physics course, we offer the basic formulas in the form of a graphic diagram as a visual cheat sheet:

The “Classic Wheel” clearly demonstrates the relationship of formulas and the interdependence of the characteristics of electric current (I - current strength, P - power, U - voltage, R - core radius)

Let us write down the dependence of the current I on the power P and line voltage U:

I = P/U_l,

Where:

• I — current strength, taken in amperes;
• P — power in watts;
• U_l — line voltage in volts.

Line voltage generally depends on the power supply source; it can be single- or three-phase.

Relationship between linear and phase voltage:

1. U_l = U*cosφ in case of single-phase voltage.
2. U_l = U*√3*cosφ in case of three-phase voltage.

For household electrical consumers, cosφ=1 is accepted, so the linear voltage can be rewritten:

1. U_l = 220 V for single-phase voltage.
2. U_l = 380 V for three-phase voltage.

Next, we summarize all consumed currents using the formula:

I = (I1+I2+…IN)*K*J,

Where:

• I – total current in amperes;
• I1..IN – current strength of each consumer in amperes;
• K – simultaneity coefficient;
• J – safety factor.

The coefficients K and J have the same values ​​as those used when calculating the total power.

There may be a case when in a three-phase network a current of unequal strength flows through different phase conductors.

This happens when single-phase and three-phase consumers are connected to a three-phase cable at the same time. For example, a three-phase machine and single-phase lighting are powered.

A natural question arises: how is the cross-section of a stranded wire calculated in such cases? The answer is simple - calculations are made based on the most loaded core.

Stage #2 - selecting a suitable section using tables

The operating rules for electrical installations (PEU) contain a number of tables for selecting the required cross-section of the cable core.

The conductivity of a conductor depends on temperature. For metal conductors, resistance increases with increasing temperature.

When a certain threshold is exceeded, the process becomes self-sustaining: the higher the resistance, the higher the temperature, the higher the resistance, etc. until the conductor burns out or causes a short circuit.

The next two tables (3 and 4) show the cross-section of conductors depending on currents and installation method.

Table 3. First, you need to choose the method of laying the wires, this determines how effective the cooling is (+)

A cable differs from a wire in that all cable cores, equipped with their own insulation, are twisted into a bundle and enclosed in a common insulating sheath. More details about the differences and types of cable products are written in this article.

Table 4. The open method is indicated for all conductor cross-section values, however, in practice, sections below 3 mm2 are not laid openly for reasons of mechanical strength (+)

When using tables, the following coefficients are applied to the permissible continuous current:

• 0.68 if 5-6 cores;
• 0.63 if 7-9 cores;
• 0.6 if 10-12 cores.

Reduction factors are applied to current values ​​from the “open” column.

The neutral and grounding conductors are not included in the number of conductors.

According to PES standards, the selection of the cross-section of the neutral conductor according to the permissible continuous current is made as at least 50% of the phase conductor.

The next two tables (5 and 6) show the dependence of the permissible long-term current when laying it in the ground.

Table 5. Dependencies of permissible long-term current for copper cables when laid in air or ground

The current load when laid openly and when laid deep into the ground differs. They are accepted as equal if laying in the ground is carried out using trays.

Table 6. Dependencies of permissible continuous current for aluminum cables when laid in air or ground

For the installation of temporary power supply lines (carrying, if for private use), the following table (7) applies.

Table 7. Permissible continuous current when using portable hose cords, portable hose and shaft cables, floodlight cables, flexible portable wires. Only copper conductors are used

When laying cables in the ground, in addition to the heat dissipation properties, it is necessary to take into account the resistivity, which is reflected in the following table (8):

Table 8. Correction factor depending on the type and resistivity of the soil for the permissible long-term current, when calculating the cable cross-section (+)

Calculation and selection of copper cores up to 6 mm² or aluminum up to 10 mm² is carried out as for continuous current.

In the case of large cross-sections, it is possible to apply a reduction factor:

0.875 * √T_pv

Where T_pv — ratio of switching duration to cycle duration.

The duration of switching on is taken to be no more than 4 minutes. In this case, the cycle should not exceed 10 minutes.

When choosing a cable for distributing electricity in wooden house Particular attention is paid to its fire resistance.

Stage #3 - calculation of the current cross-section of the conductor using an example

Task: calculate the required section copper cable for connection:

• three-phase woodworking machine with a power of 4000 W;
• three-phase welding machine with a power of 6000 W;
• household appliances in the house with a total power of 25,000 W;

The connection will be made with a five-core cable (three phase conductors, one neutral and one grounding), laid in the ground.

The insulation of cable and wire products is calculated for a specific operating voltage. It should be taken into account that the operating voltage of his product indicated by the manufacturer must be higher than the mains voltage

Solution.

Step 1. We calculate the linear voltage of a three-phase connection:

U_l = 220 * √3 = 380 V

Step #2. Household appliances, a machine tool and a welding machine have reactive power, so the power of the machinery and equipment will be:

P_those = 25000 / 0.7 = 35700 W

P_obor = 10000 / 0.7 = 14300 W

Step #3. Current required to connect household appliances:

I_those = 35700 / 220 = 162 A

Step #4. Current required to connect equipment:

I_obor = 14300 / 380 = 38 A

Step #5. The required current for connecting household appliances is calculated based on one phase. According to the problem, there are three phases. Therefore, the current can be distributed among the phases. For simplicity, we assume a uniform distribution:

I_those = 162 / 3 = 54 A

Step #6. Current per phase:

I_f = 38 + 54 = 92 A

Step #7. Equipment and household appliances will not work at the same time; in addition, we will set aside a reserve of 1.5. After applying correction factors:

I_f = 92 * 1.5 * 0.8 = 110 A

Step #8. Although the cable contains 5 cores, only three phase cores are taken into account. According to Table 8 in the column three-core cable in the ground, we find that a current of 115 A corresponds to a core cross-section of 16 mm².

Step #9. According to Table 8, we apply a correction factor depending on the characteristics of the land. For a normal type of earth, the coefficient is 1.

Step #10. Optional, calculate the diameter of the core:

D = √(4*16 / 3.14) = 4.5 mm

If the calculation were made only based on power, without taking into account the peculiarities of cable laying, then the cross-section of the core would be 25 mm². Calculating current strength is more complicated, but sometimes allows you to save significant money, especially when it comes to multi-core power cables.

You can read more about the relationship between voltage and current values here.

Voltage drop calculation

Any conductor, except superconductors, has resistance. Therefore, if the cable or wire is long enough, a voltage drop occurs.

PES standards require that the cross-section of the cable core be such that the voltage drop is no more than 5%.

Table 9. Resistivity of common metal conductors (+)

This primarily concerns low-voltage cables of small cross-section.

The voltage drop calculation is as follows:

R = 2*(ρ * L) / S,

U_pad = I * R,

U_% = (U_pad /u_lin) * 100,

Where:

• 2 – coefficient due to the fact that the current necessarily flows through two wires;
• R – conductor resistance, Ohm;
• ρ — conductor resistivity, Ohm*mm²/m;
• S – conductor cross-section, mm²;
• U_pad – drop voltage, V;
• U_% - voltage drop relative to U_lin,%.

Using formulas, you can independently perform the necessary calculations.

Carrying calculation example

Task: calculate the voltage drop for a copper wire with a cross-section of one core of 1.5 mm². The wire is required to connect a single-phase electric welding machine with a total power of 7 kW. Wire length 20 m.

Those wishing to connect a household welding machine to an electrical network branch should take into account the current sieve for which the cable used is designed. It is possible that the total power of operating devices may be higher. The best option is to connect consumers to separate branches

Solution:

Step 1. We calculate the resistance of the copper wire using Table 9:

R = 2*(0.0175 * 20) / 1.5 = 0.47 Ohm

Step #2. Current flowing through the conductor:

I = 7000 / 220 = 31.8 A

Step #3. Voltage drop on the wire:

U_pad = 31.8 * 0.47 = 14.95 V

Step #4. We calculate the percentage of voltage drop:

U_% = (14,95 / 220) * 100 = 6,8%

Conclusion: to connect the welding machine, a conductor with a large cross-section is required.

Conclusions and useful video on the topic

Calculation of the conductor cross-section using the formulas:

Recommendations from specialists on the selection of cable and wire products:

The above calculations are valid for copper and aluminum conductors for industrial use. For other types of conductors, the total heat transfer is pre-calculated.

Based on these data, the maximum current that can flow through the conductor without causing excessive heating is calculated.

If you have any questions about the method for calculating the cable cross-section or would like to share your personal experience, please leave comments on this article.The review section is located below.