When transmitting electric current, uneven operation of consumers in different parts of the circuit is possible. There can be several reasons for this phenomenon, and the main one is the voltage drop.

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Basic formulas for determining stress

To calculate the voltage and resistance in the circuit, formulas are used or ready-made online calculator s.

Through current and resistance

Ohm's law has exceptions to apply:

1. With the passage of high-frequency currents, a rapid change in electromagnetic fields occurs. When calculating high-frequency circuits, the inertia of the particles that carry the charge should be taken into account.
2. When circuits operate at low temperatures (near absolute zero), substances may develop the property of superconductivity.
3. A conductor heated by passing currents causes a variable resistance.
4. When exposed to high voltage conductors or dielectrics.
5. During the processes taking place in devices based on semiconductors.
6. When LEDs work.

Through power and current

With a known consumer power and current strength, the voltage is calculated by the formula U \u003d P / I, where P is the power in Watts, and I is the current strength in Amps.

When calculating in AC circuits, a different formula is used: U=(P/I)*cosφ, where cosφ is the power factor, depends on the nature of the load.

When using devices with an active load (incandescent lamps, devices with heating coils and elements), the coefficient approaches one. The calculations take into account the possibility of the presence of a reactive component during the operation of devices and the value of cosφ is considered equal to 0.95. When using devices with a reactive component (electric motors, transformers), it is customary to consider cosφ equal to 0.8.

Through work and charge

The calculation method is used in laboratory tasks and is not used in practice.

The formula has a form similar to Ohm's law: U = A / q, where A is the work done to move the charge in Joules, and q is the charge passed, measured in Coulomb.

Resistance calculation

During operation, the conductor creates an obstacle to the flow of electric current, which is called resistance. In electrical calculations, the concept of resistivity is used, which is measured in Ohm*m.

Serial connection

When connected in series, the output of an element is connected to the input of the next one. The total resistance is found using the calculation formula: R=R1+R2+…+Rn, where R=R1+R2+…+Rn are the resistance values ​​of the elements in Ohms.

Parallel connection

Parallel is a connection in which both terminals of one circuit element are connected to the corresponding contacts of another. Parallel connection is characterized by the same voltage on the elements. The current on each element will be proportional to the resistance.

The total resistance is calculated by the formula: 1/R=1/R1+1/R2+…+1/Rn.

In real wiring diagrams, a mixed connection is used. To calculate the resistance, simplify the circuit by summing the resistances in each series circuit. Then the circuit is reduced by calculating the individual sections of the parallel connection.

Voltage loss

Voltage loss is the consumption of electrical energy to overcome resistance and heat the wires.

Voltage drops occur during the operation of various electronic components, such as diodes. It is made up of the sum of the threshold voltage p-n junction and the current passing through the diode, multiplied by the resistance.

When current passes through a resistor, a voltage drop is also observed. This effect is used to reduce the voltage in certain sections of the circuits. For example, to use devices designed for low voltage in circuits with a high voltage value.

Series connection of resistance

The diagram shows an example of a series connection of a resistor, which causes a voltage drop across the lamp from 12 to 7 volts. Light intensity regulators (dimmers) are built on this principle.

When operating wiring with a length of up to 10 meters, voltage losses can be neglected.

The voltage loss across the resistor and measurement methods are shown in the video from the Radio Amateur TV channel.

What causes voltage loss

Voltage losses in the cable system are the causes of a number of negative phenomena:

• defective and incorrect work of consumers;
• equipment damage and failure;
• decrease in power and torque of electric motors (especially noticeable at the time of start-up);
• uneven distribution of current between consumers in the initial section and at the end of the circuit;
• Due to the operation of the lamps at an incomplete glow, there is an incomplete use of the current power, which leads to energy losses.

What does loss depend on?

Voltage loss in AC and DC voltage circuits depends on the strength of the current and the resistance of the conductor. With an increase in these parameters, voltage losses increase. In addition, the design of the cables also influences the loss. The pressing density and the degree of insulation of the conductors in the cable turn it into a capacitor, which forms a charge with a capacitance.

The voltage loss across diodes depends on the type of material. When using germanium, the value lies in the range of 0.5-0.7 volts, on cheaper silicon, the value increases and reaches 0.7-1.2 volts. In this case, the drop does not depend on the voltage in the circuit, but depends only on the strength of the current.

The main causes of current losses in highways include:

Another reason for the voltage drop on the lines is the theft of electricity.

In domestic conditions, voltage losses depend on a number of factors:

• energy costs for heating wiring due to increased consumption;
• poor contact on the connections;
• capacitive and inductive nature of the load;
• use of obsolete consumers.

The reasons for the voltage drop are outlined in the video from the ElectronicsClub channel.

Valid values

The voltage loss value refers to the regulated values ​​​​and is standardized by several rules and instructions of the PUE (Electrical Installation Rules).

According to the rule SP 31-110-2003, the total value of the voltage loss from the point of entry into the building to the most remote consumer should not exceed 7.5%. The rule applies to electrical networks with an operating voltage of not more than 400 volts. This document is taken into account during the design of networks and acceptance and verification by Rostekhnadzor specialists.

Rule SP 31-110-2003 separately stipulates the voltage deviation in domestic single-phase current networks, which should not exceed ± 5% during normal operation of the network and ± 10% in the post-emergency. When operating low-voltage networks (up to 50 Volts), a deviation of ± 10% is normal.

To account for losses in the power supply cables, instruction RD 34.20.185-94 is used, which allows losses of no more than 6% at a voltage of 10 kV and no more than 4-6% at a voltage of 380 Volts. At the same time, a lower value applies to buildings with large losses in house wiring (for example, multi-storey residential buildings with big amount porches or sections). A larger value is taken for buildings with low internal losses (low-rise buildings or high-rise buildings with one or two entrances).

For the simultaneous fulfillment of the requirements of SP 31-110-2003 and RD 34.20.185-94, it is necessary to achieve a reduction in voltage loss to the norm of 1.5% (low-rise buildings) or 2.5% (high-rise buildings). The calculation must take into account data on cables, starting from the substation and ending with the connection to the switchboard. The voltage drop is influenced by the cross section and material of the cores, the length of the wiring, and the state of the insulation.

Since the beginning of 2013, the new standard GOST R 50571.5.52-2011 came into force, among other things, regulating the voltage drop on networks up to 0.4 kV. The document states that the drop should not exceed 3% for lighting circuits and 5% for other consumers. For cable lengths over 100 meters, the voltage drop can be corrected by 0.005% per meter of excess. In this case, the maximum adjustment parameter cannot exceed 0.5%.

The document does not indicate which wiring is subject to losses - from the switchboard to the most remote consumer or from the substation to the final luminaire. When calculating networks, the standard is interpreted as relating to the voltage drop from the shield to the most distant lamp (otherwise it completely contradicts the current SP 31-110-2003 and RD 34.20.185-94).

Based on the documentation described above, designers are trying to achieve a voltage drop inside the building of no more than 3% with a loss in the section from the substation to the switchboard of no more than 4.5%. This rule applies to 220V and 380V circuits.

Formulas

One of the main parameters for calculating the drop is resistivity.

To carry out wiring from the substation to the switchboard and further through the building, a copper or aluminum wire is used, which have specific resistances:

• 0.0175 Ohm*mm2/m for copper;
• 0.0280 Ohm*mm2/m for aluminium.

• to determine the rated current that will pass through the conductor: I \u003d P / U, where P is the transmitted power (Watts), and U is the rated voltage (Volts);
• to determine the resistance: R \u003d (2 * ρ * L) / s, where ρ is the specific resistance of the conductor, s is the wire cross section (mm2), and L is the line length (mm);
• the voltage loss in the wire is: ΔU=(2*I*L)/(γ*s), where L is the line length (mm), γ is the reciprocal of the resistivity, and s is the wire cross section (mm2);
• using the formula s=(2*I*L)/(γ*ΔU), you can calculate the required wire cross-section according to the required load or perform a verification calculation of the loss.

According to the known cross section, it is possible to determine the diameter of the wire using formulas or tables, which is then compared with the real value.

The voltage drop over long sections of single-phase current networks can be calculated using the formulas:

How to determine voltage loss

In networks with voltages up to 220V, losses can be determined using a voltmeter.

1. Measure at the beginning of the chain.
2. Perform a voltage measurement at the most remote area.
3. Calculate the difference and compare with the standard value. With a large drop, it is recommended to check the condition of the wiring and replace the wires with products with a smaller cross section and resistance.

The second way is the calculation by formulas.

Calculation examples

The basic way to calculate power loss can be an online calculator that makes calculations based on the initial data (length, cross section, load, voltage and number of phases).

Loss Calculator Sample

An example of a calculation using formulas for a residential building is the task of determining the voltage drop in a single room. The maximum rated power is 4 kW at a current of 16 A, the wiring is made of an aluminum conductor with a cross section of 1.5 squares and has a length of 40 meters.

The drop will be: U \u003d (p * L * 2) / (s * I) \u003d 0.028 * 40 * 2 / 1.5 * 16 \u003d 9.33 V. The voltage, taking into account the loss, will be 220-9.33 \u003d 210.67 B (or 4.2%). The value is at the tolerance limit, there is a risk of operation of consumers with partial power (especially in the event of a drawdown of the main voltage of 220 V).

In a more detailed and accurate calculation, it is necessary to take into account the reactive and active components of the resistance and the transmitted power. An example of a complex calculation is a trunk line made using a four-wire SIP cable. Four branches are connected to the highway, to which country houses are connected. The load power factor is taken as 0.98. The main SIP2 cable has four cores of 50 mm2, the SIP4 cable for connecting a house has two cores of 16 mm2. Distances are shown on the diagram.

Wiring diagram

For the calculation you need:

1. Determine the resistance per unit length of the SIP2 wiring according to the reference book: Rpo = 0.641 10-3 Ohm / m. Xlimit=0.0794 10-3 Ohm/m.
2. Find out similar values ​​for SIP4: Rpo = 1.91 10-3 Ohm/m. Xth = 0.0754 10-3 Ohm/m
3. For a three-phase section, the calculation is carried out according to the formula: ΔU=((L*(P*Rpog+Q*Xpog))/U2)*100.
4. For single-phase branches: ΔU=((2*L*(P*Rline+Q*Xline))/U2)*100, where P and Q are the calculated active power of the line (W), L is the length of the line section (m), Rpog (Xpog) - linear resistance of the wire (Ohm / m), U - rated phase voltage of the network (V).

Since the value of Q*Xpog is an order of magnitude smaller than P Rpog, then in the calculations it is neglected and the formula is simplified to the form: ΔU=((L*P*Rpog)/U2)*100 and ΔU=((2*L*P *Rpog)/U2)*100.

The design power in each section is determined according to the tabular values ​​​​from SP 31-110-2003. When calculating the number of consumers in intermediate sections, it is necessary to sum up their number on a branch at the end of the section and on the next one.

In the example shown, there are 34 power consumers (houses) between nodes 1 and 2. Since the tables give values ​​only for 24 and 40 houses, for our case the value is calculated according to a linear graph: Р34=Р24-((34-24)/(40-24))*(Р24-Р40)=0.9- ((34-24 / (40-24)) * (0.9-0.76) = 0.81 kW / house.

Based on the obtained power value, the voltage loss in each section is calculated.

Table with frequent values

There are tables for determining the voltage loss (percentage for the transmission of one kilowatt per kilometer) depending on the core material, cross section and reactive power factor.

Below is an example table for a main aluminum wire in a three-phase transmission line.

The table shows that as the reactive power factor falls, the loss decreases. Additionally, the increase in the cross section of the conductor reduces the loss.

Another version of the table for single-phase and three-phase networks for electric motors and lighting.

For example, a three-phase motor operates at a current of 100 A and a voltage of 400 V, but at the time of starting it consumes up to 500 A. Under various operating conditions, the cosine φ will be 0.8 or 0.35. To power the engine, a wire 50 meters long with a cross section of 35 squares was laid. Under normal conditions, on a three-phase network, the losses are one volt per kilometer of wiring (from the table).

In our case, the loss will be 1v * 0.05km * 100a = 5 volts. At the time of start-up, a voltage drop of 10 V is observed on the shield. Thus, the total drop will reach 15 volts, which is 3.75%. The value lies within the tolerance of the PUE and such a circuit is applicable to operation.

Cable selection

To select a cable for heating and voltage drop, you can use ready-made online calculators.

One of the calculators

A formula calculation method is possible, but it is used when designing wiring for large residential buildings and industrial premises.

At home, we often use portable extension cords - sockets for temporary ( usually permanent) turning on household appliances: electric heater, air conditioner, iron with high consumption currents.
The cable for this extension cord is usually selected according to the principle - whatever is at hand, and this does not always correspond to the necessary electrical parameters.

Depending on the diameter (or cross-section of the wire in mm2), the wire has a certain electrical resistance for the passage of electric current.

The larger the cross section of the conductor, the lower its electrical resistance, the lower the voltage drop across it. Accordingly, the power loss in the wire for its heating is less.

Let us carry out a comparative analysis of the power loss for heating in the wire, depending on its transverse sections. Let's take the most common cables in everyday life with a cross section: 0.75; 1.5; 2.5 mm2 for two extensions with cable length: L = 5m and L = 10m.

Take for example a load in the form of a standard electric heater with electrical parameters:
- supply voltageU = 220 Vol T ;
- electric heater power P \u003d 2.2 kW \u003d 2200 W ;
- consumption current I = P / U = 2200 W / 220 V = 10 A.

From the reference literature, we take the resistance data of 1 meter of wire of different cross sections.

A table of resistances of 1 meter of wire made of copper and aluminum is given.

Let's calculate the power loss for heating for the cross section of the wire S = 0.75 mm2 The wire is made of copper.

Resistance of 1 meter wire (from the table) R 1 \u003d 0.023 Ohm.
Length of cable L=5 meters.
Wire length in cable (round trip)2 L=2 · 5 = 10 meters.
Electrical resistance of a wire in a cable R \u003d 2 L R 1 \u003d 2 5 0.023 \u003d 0.23 Ohm.

Voltage drop in the cable during the passage of current I = 10A will: U \u003d I R \u003d 10 A 0.23 Ohm \u003d 2.3 V.
The power loss for heating in the cable itself will be: P = U I = 2.3 V 10 A = 23 W.

If the cable length L = 10 m. (of the same cross-section S = 0.75 mm2), the power loss in the cable will be 46 W. This is approximately 2% of the power consumed by the electric heater from the network.

For a cable with aluminum conductors of the same section S = 0.75 sq. mm. readings increase and amount to L=5m-34.5w. For L = 10 m - 69 W.

All calculation data for cables with a cross section of 0.75; 1.5; 2.5 mm2 for cable length L=5 and L=10 meters are shown in the table.
Where: S - wire cross-section in mm.sq.;
R1- resistance of 1 meter of wire in ohms;
R is the cable resistance in ohms;
U is the voltage drop in the cable in Volts;
P is the power loss in the cable in watts or in percent.

What conclusions should be drawn from these calculations?

• - With the same cross section, the copper cable has a greater margin of safety and less electrical power loss for heating the wire R.
• - With an increase in the length of the cable, the losses P increase. To compensate for the losses, it is necessary to increase the cross-section of the cable wires S.
• - It is desirable to choose a cable in a rubber sheath, and the cable cores are stranded.

For an extension cord, it is desirable to use a Euro socket and a Euro plug. The Euro plug pins are 5 mm in diameter. A simple electrical plug has a 4 mm pin diameter. Euro plugs are rated for more current than a simple socket and plug. The larger the diameter of the plug pins, the larger the contact area. at the junction of the plug and socket,hence lower contact resistance. This contributes to less heating at the junction of the plug and socket.

In order to ensure the supply of voltage from the switchgear to the end consumer, power lines are used. They can be aerial or cable and have a significant length.

Like all conductors, they have a resistance that depends on the length and the longer they stretch, the greater the voltage loss.

And the longer the line, the greater the voltage loss. Those. the voltage at the input and at the end of the line will be different.

In order for the equipment to work without failures, these losses are normalized. They should have a total value not exceeding 9%.

The maximum voltage drop at the input is five percent, and to the most remote consumer no more than four percent. In a three-phase network with a three- or four-wire network, this figure should not exceed 10%.

If these indicators are not met, end users will not be able to meet the nominal parameters. With a decrease in voltage, the following symptoms occur:

• Lighting devices that use incandescent lamps begin to work (glow) at half incandescence;
• When the electric motors are turned on, the starting force on the shaft decreases. As a result, the motor does not rotate, and as a result, the windings overheat and fail;
• Some electrical appliances do not turn on. There is not enough voltage, and other devices after switching on can fail;
• Installations that are sensitive to input voltage are unstable, and light sources that do not have an incandescent filament may not turn on.

Electricity is transmitted via air or cable networks. Air ones are made of aluminum, and cable ones can be aluminum or copper.

In cables, in addition to active resistance, there is capacitive resistance. Therefore, the power loss depends on the length of the cable.

Causes leading to a decrease in voltage

Voltage losses in power lines occur for the following reasons:

• A current passes through the wire, which heats it, as a result, active and capacitive resistance increases;
• A three-phase cable with a symmetrical load has the same voltage values ​​\u200b\u200bin the cores, and the current of the neutral wire will tend to zero. This is true if the load is constant and purely active, which is impossible in real conditions;
• In networks, in addition to the active load, there is a reactive load in the form of transformer windings, reactors, etc. and as a consequence, inductive power appears in them;
• As a result, the resistance will be composed of active, capacitive and inductive. It affects the voltage losses in the network.

The current loss depends on the cable length. The longer it is, the greater the resistance, which means that the losses are more significant. It follows that the power loss in the cable depends on the length or length of the line.

Loss value calculation

To ensure the operability of the equipment, it is necessary to make a calculation. It is carried out at the time of design. The current level of development of computer technology allows you to make calculations using an online calculator that allows you to quickly calculate cable power losses.

To calculate, just enter the necessary data. Set the parameters of the current - direct or alternating. The transmission line material is aluminum or copper. They indicate by what parameters the power loss is calculated - by the cross section or diameter of the wire, load current or resistance.

In addition, they indicate the mains voltage and cable temperature (depending on the operating conditions and the method of laying). These values ​​are substituted into the calculation table and calculated using an electronic calculator.

You can calculate based on mathematical formulas. In order to correctly understand and evaluate the processes occurring during the transmission of electrical energy, a vector form of representation of characteristics is used.

And to minimize calculations, a three-phase network is presented as three single-phase networks. Network resistance is represented as a series connection of active and reactive resistance to the load resistance.

In this case, the formula for calculating the power loss in the cable is greatly simplified. To obtain the necessary parameters, use the formula.

This formula shows the power loss of a cable as a function of current and resistance distributed along the length of the cable.

However, this formula is valid if you know the current strength and resistance. Resistance can be calculated using the formula. For copper, it will be equal to p = 0.0175 Ohm * mm2 / m, and for aluminum p = 0.028 Ohm * mm2 / m.

Knowing the value of resistivity, the resistance is calculated, which will be determined by the formula

R \u003d p * I / S, where p is the resistivity, I is the length of the line, S is the cross-sectional area of ​​\u200b\u200bthe wire.

In order to calculate the voltage loss along the length of the cable, it is necessary to substitute the obtained values ​​​​into the formula and perform calculations. These calculations can be made during the installation of electrical networks or security systems and video surveillance.

If power loss calculations are not performed, this can lead to a decrease in the supply voltage of consumers. As a result, the cable will overheat, it can get very hot, and as a result, the insulation will be damaged.

This may result in electric shock or short circuit to persons. A decrease in voltage in the line can lead to the failure of their electronic equipment.

Therefore, when designing electrical wiring, it is important to calculate the voltage loss in the supply wires and the laid cable.

Loss reduction methods

Power losses can be reduced by the following methods:

• Increase the cross section of the conductors. As a result, resistance will decrease, and losses will decrease;
• Reducing power consumption. This setting cannot always be changed;
• Changing the length of the cable.

Reducing the power and changing the length of the line is practically impossible. Therefore, if you increase the wire cross section without calculation, then on a long line this will lead to unjustified costs.

And this means that it is very important to make a calculation that will allow you to correctly calculate the power losses in the cable and choose the optimal value for the cross section of the cores.

Interested in normalizing the loss of voltage in the lines in various sections of the electrical network:

CPU - TP (RTP) - ASU (GRSHCH) - SCHO (SCHR or SC) - n.a. EO lamp (the most powerful n.o. EP).

Accepted abbreviations (for definitions, see chapter 7.1 of the EMP and at the end of this article):

• feasibility study - feasibility study,
• CPU - power center,
• TP - transformer substation,
• RTP - distribution transformer substation,
• ASU - input-distribution device,
• Main switchboard - main switchboard,
• SCHO - working lighting shield,
• ShCHAO - emergency lighting panel,
• ShchR - switchboard,
• ShchS - power shield,
• EO - electric lighting,
• EP - electrical receiver,
• EU - electrical installation,
• well. - the most distant
• r.l. - distribution line
• gr.l. - group line
• d.c.o.s. – admissible values ​​of steady-state voltage deviation.

Voltage loss in the power supply system is a value equal to the difference between the steady-state values ​​of the effective voltage measured at two points of the power supply system (GOST 23875-88 "Quality of electrical energy. Terms and definitions"), for example, the algebraic difference between the voltage at the beginning (for example, at the source supply) and at the end (at the terminals of the power receiver) of the line.

On the secondary windings of TP transformers, the voltage is 0.4 kV (clause 1.2.23 of the PUE of the 7th edition), i.e. 105% of the rated voltage of the electrical network 0.38 kV (GOST 721 and GOST 21128). We have a “disposable” voltage loss in normal mode from the TP busbars to the ASP - the average value is within 4-6% (clause 5.2.4 of RD 34.20.185-94). Normally permissible values ​​of the steady-state voltage deviation at the EA terminals are ± 5% of the rated mains voltage (clause 5.2 of GOST 13109-97).

We have a “disposable” voltage loss of ≈10% from the busbars of the switchgear 0.4 kV TS to n.o. ES, but it is recommended that the total voltage losses from the busbars of the transformer substation to n.o. EO lamps did not exceed 7.5% (SP 31-110-2003). So, if from the busbars 0.4kV TS to the ASU - 5%, then in the section from the ASU to the n.s. EE lamps are not more than 2.5%, and for the rest of the ED, the losses in the building EE should not exceed 4% (GOST R 50571.15-97):

• from TP tires to ASU - 5% (380V);
• from tires TP to n.o. EO lamps - 7.5% (370V);
• from tires TP to n.o. EP - 9% (364.8V).

And voltage losses in the power plant of the building in various sections of the electrical network, i.e. r.l. and gr.l. (see columns "b" and "c" of Table 1), are not standardized and are selected based on specific conditions, feasibility studies, etc. From the point of view of reducing the complexity of design, voltage losses in various sections of the electrical network, in my opinion, can be taken as follows, from ASU to:

• well. EO lamps not more than 2.5%, of which
• r.l. up to SCHO - 0.5%,
• gr.l. BC EO lamps - 2%.
• well. EP should not exceed 4%, of which
• r.l. up to SHR - 2%,
• line BC EP - 2%.
• electric motor, electronic equipment and special equipment - according to the passport, but not more than 15%.
• For voltage circuits of electricity meters - 0.5% (RM-2559).

It is not required to calculate the voltage loss in each group line (with equal conductor cross-sections) in the networks of internal EE and socket outlets, because there are no current guidelines obliging to make such a calculation, which is only necessary to determine the values ​​under the worst conditions, i.e. for n.s. EO lamps and the most loaded n.s. line EP.

According to the experience of designing, voltage losses in intra-apartment group lines of general lighting can be taken equal to 1-0.8% (Tulchin I.K., Nudler G.I., Electrical networks and electrical equipment of residential and public buildings - 2nd ed., M. : Energoatomizdat, 1990, see Table 16.1 "Limits of permissible voltage losses at which the parameters of the electrical network have values ​​close to optimal" on page 253).

On tires n / n TP during the period of the lowest loads of networks is not higher than 100% of the rated voltage (clause 1.2.23 of the Electrical Installation Code of the 7th edition) and voltage losses, depending on the load power in the networks, decrease proportionally.

But that is not all! It is necessary to make a calculation for voltage losses in the post-emergency mode in order not to go beyond the maximum permissible values ​​​​of the steady-state voltage deviation (GOST 13109-97): ± 10% of the rated voltage of the electrical network according to GOST 721 and GOST 21128 (nominal voltage). Calculation for voltage losses in the post-accident mode can be. relevant, for example, for mutually redundant cable lines.

Position of Rostekhnadzor:
Information and reference publication “Electrical Engineering News”,
annual supplement "Question-Answer", supplement to the journal No. 6(48) 2007.

There are many disagreements among designers in the understanding of SP 31-110-2003, clause 7.23. Deviation of voltage from the nominal voltage at the terminals of power receivers and n.o. EO lamps should not exceed 5% in norms. mode, and from tires TP to n.s. EO lamps - 7.5%. So, ASU - n.s. EO lamps - 5% of 380/220 V, but then it is necessary to apply increased voltage from the transformer substation to the ASU in order to obtain the nominal voltage value in the ASU, taking into account the losses on this line (2.5%).

First of all, it is necessary to separate the concepts of "voltage deviation" and "voltage loss". In the first paragraph of clause 7.23 of SP 31-110-2003, the voltage deviation from the nominal voltage at the terminals of electrical receivers of incandescent lamps is normalized. In the third paragraph of clause 7.23 of SP 31-110-2003, we are talking about the loss of voltage in the lines in the section from the 0.4 kV buses of the 6-10 / 0.4 kV transformer substation to the most remote electrical receiver.
Compliance with the condition of the first paragraph is mandatory, the third paragraph is recommended.
In accordance with the instructions of clause 1.2.23 of the EMP 7th edition, the voltage on the 3–20 kV busbars of power plants and substations must be maintained at least 105% of the nominal during the period of the highest loads and at least 100% of the nominal during the period of the lowest loads in these networks.
Taking into account these initial provisions, it is necessary to check the sections of conductors selected according to other conditions. The voltage loss in the lines in normal mode should be such that the voltage at the terminals of the most remote power receiver, both at the highest and at the lowest loads, is within ± 5% of the nominal. When checking the sections of selected conductors for voltage loss, it is necessary to take into account the position of the tap changer at transformer substations with a voltage of 6–10/0.4 kV.

Viktor Shatrov, referent of Rostekhnadzor.

Normative references:

PUE 7th edition.
Voltage levels and regulation, reactive power compensation.

1.2.22. For electrical networks, technical measures should be provided to ensure quality of electrical energy in accordance with the requirements of GOST 13109.

1.2.23. Voltage regulation devices must ensure that the voltage on the buses with a voltage of 3-20 kV of power plants and substations to which distribution networks are connected is not lower than 105% of the nominal during the period of the greatest loads and not higher than 100% of the nominal during the period of the least loads of these networks. Deviations from the specified voltage levels must be justified.

1.2.24. The choice and placement of reactive power compensation devices in electric networks are based on the need to ensure the required network throughput in normal and post-accident modes while maintaining the required voltage levels and stability margins.

GOST 13109-97. Standards for the quality of electrical energy in general-purpose power systems.5.2. Voltage deviation.

The voltage deviation is characterized by an indicator of the steady-state voltage deviation, for which has the following standards:

• normally permissible and maximum permissible values ​​of the steady-state voltage deviation δUу at the outputs of electrical energy receivers are respectively ± 5 and ± 10% of the rated voltage of the electrical network according to GOST 721 and GOST 21128 (rated voltage);
• normally allowable and maximum allowable values ​​of the steady voltage deviation at the points of general connection of consumers of electrical energy to electrical networks with a voltage of 0.38 kV or more should be established in contracts for the use of electrical energy between the power supply organization and the consumer, taking into account the need to comply with the norms of this standard at the terminals of the receivers electrical energy.

RD 34.20.185-94
Instructions for the design of urban electrical networks.
Ch. 5.2 Voltage levels and regulation, reactive power compensation

5.2.4. Preliminary selection of cross-sections of wires and cables is allowed to be made based on the average values ​​of the maximum voltage losses in normal mode: in networks of 10 (6) kV no more than 6%, in networks of 0.38 kV (from transformer substations to inputs to buildings) no more than 4-6 %.

Larger values ​​refer to lines supplying buildings with a lower voltage loss in intra-house networks (low-rise and one-section buildings), smaller values ​​- to lines supplying buildings with a greater voltage loss in intra-house networks (multi-storey multi-section residential buildings, large public buildings and institutions).

SP 31-110-2003
Design and installation of electrical installations of residential and public buildings.
7. Schemes of electrical networks.

7.23 Voltage deviations from the rated voltage at the terminals of power receivers and the most distant electric lighting lamps should not exceed ± 5% in normal mode, and the maximum allowable in post-emergency mode at the highest design loads - ± 10%. In networks with a voltage of 12-50 V (counting from a power source, for example, a step-down transformer), voltage deviations can be accepted up to 10%.

For a number of electrical receivers (control devices, electric motors), voltage reduction in starting modes is allowed within the limits of the values ​​​​regulated for these electrical receivers, but not more than 15%.

Taking into account the regulated deviations from the nominal value, the total voltage losses from the 0.4 kV TS buses to the most remote general lighting lamp in residential and public buildings are not should generally exceed 7.5%. The range of voltage changes at the terminals of electrical receivers when starting the electric motor should not exceed the values ​​established by GOST 13109.

GOST R 50571.15-97 (IEC 364-5-52-93). Electrical installations of buildings.
Part 5. Selection and installation of electrical equipment. Chapter 52
525. Loss of voltage in electrical installations of buildings.

Voltage losses in electrical installations of buildings should not exceed 4% of the rated voltage of the installation. Temporary conditions such as transients and voltage fluctuations [caused by incorrect (erroneous) switching] are not taken into account.

IEC 60364-7-714-1996, IEC 60364-7-714 (1996). Electrical installations of buildings.
Part 7: Requirements for special installations or rooms.
Section 714 Outdoor lighting installations.

714.512. The voltage drop under normal operating conditions must be compatible with the conditions resulting from the inrush current of the lamps.

RD 34.20.501-95
Rules for the technical operation of power stations and networks of the Russian Federation.
5. Electrical equipment of power plants and networks.

5.12.7. The lighting network of power plants must be powered through stabilizers or from separate transformers that ensure the ability to maintain the lighting voltage within the required limits. The voltage on the lamps should not be higher than the rated voltage. The voltage drop at the most remote lamps of the internal working lighting network, as well as searchlight installations, should be no more than 5% of the rated voltage; for the most remote lamps of the outdoor and emergency lighting network and in the 12-42 V network, no more than 10% (for fluorescent lamps, no more than 7.5%).

GOST R IEC 60204-1-99 (IEC 60204-1). Machine safety.
Electrical equipment of machines and mechanisms. General requirements.
13 Cables and wires. 13.5 Voltage drop on wires

Under normal operating conditions, the voltage drop in the section from the power source to the point of application of the load should not exceed 5% of the nominal voltage.

RM 2559
Instructions for the design of electricity consumption metering in residential and public buildings.

5.15. The cross section and length of the wires and cables used for the voltage circuits of the meters must be selected so that the voltage loss is not more than 0.5% of the rated voltage.

Voltage losses in the cable are a big problem in the case of a long path from the power source to the consumer, as well as a high power consumption of the latter. Incorrectly selected materials for the electrical line (wiring), for example, wires with very thin strands, begin to heat up due to low conductivity for electric current. The calculator provided by us allows you to calculate the voltage losses in the cable online:

Let's also understand where the losses come from and why. Conductors are made of copper and aluminum, although they are excellent conductors, they still have a certain resistivity, which is active. A certain number of Volts falls on any resistive element, according to:

U=I*Rwire

In direct current, when calculating voltage losses in the cable, only the active resistance R appears. At the same time, when working with alternating current, for example, in 0.4 kV networks, the reactive part is added to the active value - they make up the total resistance Z (Xl and Xc) . The role of reactive power is very important in the calculations, as it is 20 percent or more of the power consumed.

Why is such a calculation needed? Everything is very simple: the more R wires - the more losses, and the more the wires heat up. Let's figure out how to calculate them manually, but it's easier to do it using an online calculator. The formula for determining the resistance of a conductor looks like this:

• p is the resistivity;
• L is the length;
• S is the cross-sectional area.

It follows that it depends on the length and cross-sectional area. The longer and thinner the conductor, the greater R, and to reduce it, conductors with a large cross section are needed.

Then, in the simplest case, the losses are equal to the voltage drop on the line:

dU=I*Rwire

And taking into account the total power for alternating current:

But the first formula is valid only for one of the conductive cores, and electricity, as you know, cannot be transmitted through one wire. It is transmitted over at least two, in a three-phase network - over four wires.

• length;
• cross-sectional area of ​​conductive conductors;
• the amount of current or power consumed;
• number of phases;
• conductor temperature;
• COS F.

As a result, in a couple of clicks, the online calculator will provide you with the following data:

• losses;
• cable resistance;
• reactive power;
• load voltage.

materials

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