When designing electrical networks with low currents, voltage losses in conductors are often calculated. The results obtained are then used to determine the optimal cross-section of current-carrying conductors. If a mistake is made during the selection of wires and cables, the electrical system will quickly fail or will not start at all. To carry out the necessary calculations, special formulas or online calculators are used.
Reasons for losses
Every electrician knows that cables are made up of cores. They are made of copper or aluminum and covered with an insulating layer. To protect against mechanical damage, the conductors are placed in an additional polymer sheath. Since the current-carrying conductors are densely located and compressed by a protective coating, when the line is long, they begin to work on the principle of a capacitor. To put it simply, a charge is created in the cores that has capacitive reactance.
The voltage loss diagram in the wires looks like this:
If this process is represented graphically, then the segment AD will be an indicator of losses.
Performing such calculations manually is quite difficult and an online calculator is now often used. Voltage losses calculated with its help turn out to be quite accurate, and the error is minimal.
Consequences of voltage reduction
In accordance with regulatory documentation, losses on the main line from the transformer to the most remote point for public facilities should not exceed 9%. As for possible losses at the point where the line enters the end user, this figure should be no more than 4%.
In case of deviation from the specified limits, the following consequences are possible:
- Volatile equipment will not be able to function normally.
- If the input voltage is low, electrical appliances may malfunction.
- The current load will not be distributed evenly between consumers.
High demands are placed on the characteristics of power lines. When designing them, it is necessary to calculate possible losses not only in the main networks, but also in the secondary ones.
Several methods can be used to calculate voltage losses. It’s worth considering everything so that every electrician can choose the most attractive one depending on the situation.
Using tables and formulas
In practice, when installing electric mains, copper or aluminum conductors are used. Knowing the resistivity of these materials, as well as the current strength and wire resistance, you can use the following voltage drop formulas:
A home master and even a specialist can use special tables. This is a fairly convenient and simple way to carry out the necessary calculations. However, in some cases it is necessary to obtain the most reliable result, taking into account the indicators of active and reactance. In such a situation, you have to use a more complex formula:
To ensure optimal load in a three-phase network, each phase must be loaded evenly. To solve this problem, electric motors should be connected to linear conductors, and lamps should be connected between the neutral line and phases.
Online services
The use of formulas, graphs and tables is a rather labor-intensive process. It is not always necessary to get the most accurate results, and in such a situation it is worth using online calculators. These services work as follows:
- Current indicators, conductor material, cross-section of current-carrying conductors and line length are entered into the program.
- You will also need to provide information on the number of phases, network voltage, power and line temperature during operation.
- After entering all the necessary data, the program will automatically perform all the necessary calculations.
At the preliminary design stage, it is worth using several services and then determining the average value. It should be recognized that there is a certain error in calculations when using online calculators.
Reduce losses
It is quite obvious that losses depend on the length of the conductor in the line. The higher this parameter is, the more the voltage drops. Several methods can be used to reduce losses:
The latter method works great in power networks that have several backup lines. It should also be remembered that the voltage may drop as the temperature of the cable increases. If additional thermal insulation measures are used during cable installation, losses can be reduced.
In the energy industry, calculating the voltage drop on the main line is one of the most important tasks. If all calculations were carried out correctly, then the consumer will not have problems with the operation of electrical equipment.
When calculating electricity losses in a cable, it is important to take into account its length, core cross-sections, inductive reactance, and wire connections. Thanks to this background information, you can independently calculate the voltage drop.
Types and structure of losses
Even the most efficient power supply systems have some actual loss of electricity. Losses mean the difference between the electrical energy given to users and what actually came to them. This is due to the imperfection of the systems and the physical properties of the materials from which they are made.
The most common type of electricity loss in electrical networks is associated with voltage loss from the length of the cable. To standardize financial expenses and calculate their actual value, the following classification was developed:
- Technical factor. It is associated with the characteristics of physical processes and can change under the influence of loads, conditional fixed costs and climatic circumstances.
- Costs of using additional supplies and providing the necessary conditions for the activities of technical personnel.
- Commercial factor. This group includes deviations due to imperfection of control and measuring instruments and other points that provoke underestimation of electrical energy.
The main causes of voltage loss
The main reason for power loss in a cable is losses in power lines. At a distance from the power plant to consumers, not only the power of electricity is dissipated, but also the voltage drops (which, if it reaches a value less than the minimum permissible, can provoke not only inefficient operation of devices, but also their complete inoperability.
Also, losses in electrical networks can be caused by the reactive component of a section of an electrical circuit, that is, the presence of any inductive elements in these sections (this can be communication coils and circuits, transformers, low and high frequency chokes, electric motors).
Ways to reduce losses in electrical networks
A network user cannot influence losses in power lines, but can reduce the voltage drop on a section of the circuit by correctly connecting its elements.
It is better to connect a copper cable to a copper cable, and an aluminum cable to an aluminum cable. It is better to minimize the number of wire connections where the core material changes, since in such places not only energy is dissipated, but also heat generation increases, which can be a fire hazard if the level of thermal insulation is insufficient. Considering the conductivity and resistivity of copper and aluminum, it is more energy efficient to use copper.
If possible, when planning an electrical circuit, it is better to connect any inductive elements such as coils (L), transformers and electric motors in parallel, since according to the laws of physics, the total inductance of such a circuit decreases, and when connected in series, on the contrary, it increases.
To smooth out the reactive component, capacitor units (or RC filters in combination with resistors) are used.
Depending on the principle of connecting capacitors and the consumer, there are several types of compensation: personal, group and general.
- With personal compensation, capacitors are connected directly to the place where reactive power appears, that is, their own capacitor is connected to an asynchronous motor, another one is to a gas-discharge lamp, another one is to a welding one, another one is for a transformer, etc. At this point, the incoming cables are relieved of reactive currents to the individual user.
- Group compensation involves connecting one or more capacitors to several elements with large inductive characteristics. In this situation, the regular simultaneous activity of several consumers is associated with the transfer of total reactive energy between loads and capacitors. The line that supplies electrical energy to a group of loads will be unloaded.
- General compensation involves inserting capacitors with a regulator in the main panel, or main switchboard. It makes an assessment based on the current consumption of reactive power and quickly connects and disconnects the required number of capacitors. As a result, the total power taken from the network is reduced to a minimum in accordance with the instantaneous value of the required reactive power.
- All reactive power compensation installations include a pair of capacitor branches, a pair of stages, which are formed specifically for the electrical network, depending on the potential loads. Typical step dimensions: 5; 10; 20; thirty; 50; 7.5; 12.5; 25 sq.
To acquire large steps (100 or more kvar), small ones are connected in parallel. Network loads are reduced, switching currents and their interference are reduced. In networks with many high harmonics of the mains voltage, capacitors are protected by chokes.
Automatic compensators provide the network equipped with them with the following advantages:
- reduce the load on transformers;
- simplify the requirements for cable cross-sections;
- make it possible to load the power grid more than is possible without compensation;
- eliminate the causes of a decrease in network voltage, even when the load is connected by long cables;
- increase the efficiency of mobile fuel generators;
- simplify starting electric motors;
- increase cosine phi;
- eliminate reactive power from circuits;
- protect against overvoltage;
- improve the regulation of network characteristics.
Cable voltage loss calculator
For any cable, voltage loss can be calculated online. Below is an online voltage cable loss calculator.
The calculator is under development and will be available soon.
Calculation using formula
ΔU, % = (Un -- U) * 100/ Un,
From this we can derive a formula for calculating electricity losses:
ΔP, % = (Un -- U) * I * 100/ Un,
where Un is the rated voltage at the network input;
I -- actual network current;
U -- voltage at a separate network element (losses are calculated as a percentage of the nominal value available at the voltage input).
Table of voltage loss along cable length
Below are the approximate voltage drops along the cable length (Knorring table). We determine the required section and look at the value in the corresponding column.
| ΔU, % | Load moment for copper conductors, kW∙m, two-wire lines for voltage 220 V | |||||
|---|---|---|---|---|---|---|
| With a conductor cross-section s, mm² equal to | ||||||
| 1,5 | 2,5 | 4 | 6 | 10 | 16 | |
| 1 | 18 | 30 | 48 | 72 | 120 | 192 |
| 2 | 36 | 60 | 96 | 144 | 240 | 384 |
| 3 | 54 | 90 | 144 | 216 | 360 | 576 |
| 4 | 72 | 120 | 192 | 288 | 480 | 768 |
| 5 | 90 | 150 | 240 | 360 | 600 | 960 |
Wire strands emit heat when current flows. The size of the current, together with the resistance of the cores, determines the degree of loss. If you have data on the cable resistance and the amount of current passing through them, you can find out the amount of losses in the circuit.
The tables do not take into account inductive reactance because when using wires, it is excessively small and cannot be equal to active.
Who pays for electricity losses
Electricity losses during transmission (if transmitted over long distances) can be significant. This affects the financial side of the issue. The reactive component is taken into account when determining the general tariff for using the rated current for the population.
For single-phase lines it is already included in the price, taking into account the network parameters. For legal entities, this component is calculated regardless of active loads and is indicated separately in the invoice provided, at a special rate (cheaper than active). This is done due to the presence of a large number of induction mechanisms (for example, electric motors) at enterprises.
Energy supervision authorities establish the permissible voltage drop, or loss standard, in electrical networks. The user pays for transmission losses. Therefore, from the consumer's point of view, it is economically beneficial to consider reducing them by changing the characteristics of the electrical circuit.
Selecting wire cross-section for direct current. Voltage drop (explanations in the article)
They say that at one time there was a rivalry between Edison and Tesla - which current should be chosen for transmission over long distances - alternating or direct? Edison was in favor of using direct current to transmit electricity. Tesla argued that alternating current was easier to transmit and transform.
Subsequently, as is known, Tesla won. Nowadays alternating current is used everywhere, in Russia with a frequency of 50 Hz. It is cheaper to transmit such current over long distances. Although, there are also DC power lines for special applications.
And if you use high voltages (for example, 110 or 10 kV), then there will be significant savings on wires compared to low voltage. I talk about this in an article about the difference between 380V and 220V.
By the way, if you are interested in what I write about, subscribe to receive new articles and join the group on VK!
Looking ahead, I will say that the calculation of the wire cross-section for direct current is based on two criteria:
- Voltage drop (loss)
- Wire heating
The first point for direct current is the most important, and the second only follows from the first.
Now in detail, in order, for those who want to UNDERSTAND.
Voltage drop on the wire
The article will be specific, with theoretical calculations and formulas. For those who are not interested in what comes from and why, I advise you to go straight to Table 2 - Selecting the wire cross-section depending on the current and voltage drop.
And also - calculation of voltage losses on a long powerful three-phase cable line. An example of calculating a real line.
So, if we take the power constant, then as the voltage decreases, the current should increase, according to the formula:
In this case, the voltage drop on the wire (losses in the wires) due to resistance is calculated based on Ohm’s law:
From these two formulas it is clear that as the supply voltage decreases, the losses on the wire increase. Therefore, the lower the supply voltage, the larger the cross-section of the wire must be used to transmit the same power.
For direct current, where low voltage is used, you have to carefully approach the issue of cross-section and length, since these two parameters determine how many volts will be wasted.
DC resistance of copper wire
The wire resistance depends on the resistivity ρ, which is measured in Ohm mm²/m. The resistivity value determines the resistance of a piece of wire 1 m long and 1 mm² cross-section.
The resistance of the same piece of copper wire 1 m long is calculated by the formula:
R = (ρ l) / S, where (3)
R – wire resistance, Ohm,
ρ – wire resistivity, Ohm mm²/m,
l – wire length, m,
S – cross-sectional area, mm².
The resistance of the copper wire is 0.0175 Ohm mm²/m; we will use this value in further calculations.
It is not a fact that copper cable manufacturers use pure copper “0.0175 standard”, so in practice the cross-section is always taken with a margin, and against wire overload they use protective circuit breakers, also with a margin.
From formula (3) it follows that for a piece of copper wire with a cross-section of 1 mm² and a length of 1 m, the resistance will be 0.0175 Ohm. For a length of 1 km – 17.5 Ohms. But this is only a theory, in practice everything is worse.
Below I will give a plate calculated using formula (3), which shows the resistance of a copper wire for different cross-sectional areas.
Table 0. Resistance of copper wire depending on cross-sectional area
Calculation of voltage drop on a wire for direct current
Now, using formula (2), we calculate the voltage drop on the wire:
U = ((ρ l) / S) I , (4)
That is, this is the voltage that will drop on a wire of a given cross-section and length at a certain current.
These are the tabular data for a length of 1 m and a current of 1A:
Table 1. Voltage drop on a 1 m copper wire of different cross-sections and a current of 1A:
This table is not very informative; it is more convenient to know the voltage drop for different currents and cross sections. Let me remind you that calculations for choosing the wire cross-section for direct current are carried out according to formula (4).
Table 2. Voltage drop for different wire cross-sections (top row) and current (left column).
Length = 1 meter
| S,mm² | 1 | 1,5 | 2,5 | 4 | 6 | 10 | 16 | 25 |
| 1 | 0,0175 | 0,0117 | 0,0070 | 0,0044 | 0,0029 | 0,0018 | 0,0011 | 0,0007 |
| 2 | 0,0350 | 0,0233 | 0,0140 | 0,0088 | 0,0058 | 0,0035 | 0,0022 | 0,0014 |
| 3 | 0,0525 | 0,0350 | 0,0210 | 0,0131 | 0,0088 | 0,0053 | 0,0033 | 0,0021 |
| 4 | 0,0700 | 0,0467 | 0,0280 | 0,0175 | 0,0117 | 0,0070 | 0,0044 | 0,0028 |
| 5 | 0,0875 | 0,0583 | 0,0350 | 0,0219 | 0,0146 | 0,0088 | 0,0055 | 0,0035 |
| 6 | 0,1050 | 0,0700 | 0,0420 | 0,0263 | 0,0175 | 0,0105 | 0,0066 | 0,0042 |
| 7 | 0,1225 | 0,0817 | 0,0490 | 0,0306 | 0,0204 | 0,0123 | 0,0077 | 0,0049 |
| 8 | 0,1400 | 0,0933 | 0,0560 | 0,0350 | 0,0233 | 0,0140 | 0,0088 | 0,0056 |
| 9 | 0,1575 | 0,1050 | 0,0630 | 0,0394 | 0,0263 | 0,0158 | 0,0098 | 0,0063 |
| 10 | 0,1750 | 0,1167 | 0,0700 | 0,0438 | 0,0292 | 0,0175 | 0,0109 | 0,0070 |
| 15 | 0,2625 | 0,1750 | 0,1050 | 0,0656 | 0,0438 | 0,0263 | 0,0164 | 0,0105 |
| 20 | 0,3500 | 0,2333 | 0,1400 | 0,0875 | 0,0583 | 0,0350 | 0,0219 | 0,0140 |
| 25 | 0,4375 | 0,2917 | 0,1750 | 0,1094 | 0,0729 | 0,0438 | 0,0273 | 0,0175 |
| 30 | 0,5250 | 0,3500 | 0,2100 | 0,1313 | 0,0875 | 0,0525 | 0,0328 | 0,0210 |
| 35 | 0,6125 | 0,4083 | 0,2450 | 0,1531 | 0,1021 | 0,0613 | 0,0383 | 0,0245 |
| 50 | 0,8750 | 0,5833 | 0,3500 | 0,2188 | 0,1458 | 0,0875 | 0,0547 | 0,0350 |
| 100 | 1,7500 | 1,1667 | 0,7000 | 0,4375 | 0,2917 | 0,1750 | 0,1094 | 0,0700 |
What explanations can be made for this table?
1. I marked in red those cases when the wire will overheat, that is, the current will be higher than the maximum permissible for a given cross-section. I used the table given on SamElektrika: Selecting the cross-sectional area of the wire.
2. Blue color - when the use of too thick wire is economically and technically impractical and expensive. The threshold was taken to be a drop of less than 1 V over a length of 100 m.
How to use the section selection table?
Table 2 is very easy to use. For example, you need to power a certain device with a current of 10A and a constant voltage of 12V. The line length is 5 m. At the output of the power supply we can set the voltage to 12.5 V, therefore, the maximum drop is 0.5 V.
Available - wire with a cross-section of 1.5 square. What do we see from the table? At 5 meters with a current of 10 A we will lose 0.1167 V x 5m = 0.58 V. It seems to be suitable, considering that most consumers tolerate a deviation of +-10%.
But. After all, we actually have two wires, plus and minus, these two wires form a cable on which the load supply voltage drops. And since the total length is 10 meters, the drop will actually be 0.58 + 0.58 = 1.16 V.
In other words, in this situation, the output of the power supply is 12.5 Volts, and the input of the device is 11.34. This example is relevant for powering an LED strip.
And this is without taking into account the contact resistance of the contacts and the imperfection of the wire (“sample” of copper is not the same, impurities, etc.)
Therefore, such a piece of cable most likely will not work; you need a wire with a cross-section of 2.5 square. It will give a drop of 0.7V on a 10m line, which is acceptable.
What if there is no other wire? There are two ways to reduce voltage loss in wires.
1. The 12.5 V power supply should be placed as close to the load as possible. If we take the example above, 5 meters will suit us. This is what they always do to save on wires.
2. Increase the output voltage of the power supply. The risk is that as the load current decreases, the voltage across the load may rise to unacceptable limits.
For example, in the private sector, 250-260 Volts are installed at the output of the transformer (substation); in houses near the substation, the light bulbs burn like candles. I mean, not for long. And residents on the outskirts of the area complain that the voltage is unstable and drops to 150-160 Volts. Loss of 100 Volts! Multiplied by the current, you can calculate the power that heats the street, and who pays for it? We, the “losses” column on the receipt.
Conclusion on choosing the wire cross-section for constant voltage:
The shorter and thicker the wire through which direct current flows, the lower the voltage drop across it, the better. That is, the voltage loss in the wires is minimal.
If you look at Table 2, you need to select the values from the top-right, without going into the “blue” zone.
For alternating current the situation is the same, but the issue is not so acute - there power is transferred by increasing the voltage and decreasing the current. See formula (1).
In conclusion, here is a table in which the DC voltage drop is set to a limit of 2%, and the supply voltage is 12 V. The required parameter is the maximum wire length.
Attention! This refers to a two-wire line, for example a cable containing 2 wires. That is, the case when, through a cable 1 m long, the current makes a path of 2 m, back and forth. I brought this option because... it is most often encountered in practice. For one wire, to find out the voltage drop across it, you need to multiply the number inside the table by 2. Thanks to attentive readers!
Table 3. Maximum wire length for 2% DC voltage drop.
| S,mm² | 1 | 1,5 | 2,5 | 4 | 6 | 10 | 16 | 25 | 35 | 50 | 75 | 100 |
| 1 | 7 | 10,91 | 17,65 | 28,57 | 42,86 | 70,6 | 109,1 | 176,5 | 244,9 | – | – | – |
| 2 | 3,53 | 5,45 | 8,82 | 14,29 | 21,4 | 35,3 | 54,5 | 88,2 | 122,4 | 171,4 | – | – |
| 4 | 1,76 | 2,73 | 4,41 | 7,14 | 10,7 | 17,6 | 27,3 | 44,1 | 61,2 | 85,7 | 130,4 | – |
| 6 | 1,18 | 1,82 | 2,94 | 4,76 | 7,1 | 11,7 | 18,2 | 29,4 | 40,8 | 57,1 | 87 | 117,6 |
| 8 | 0,88 | 1,36 | 2,2 | 3,57 | 5,4 | 8,8 | 13,6 | 22 | 30,6 | 42,9 | 65,25 | 88,2 |
| 10 | 0,71 | 1 | 1,76 | 2,86 | 4,3 | 7,1 | 10,9 | 17,7 | 24,5 | 34,3 | 52,2 | 70,6 |
| 15 | – | 0,73 | 1,18 | 1,9 | 2,9 | 4,7 | 7,3 | 11,8 | 16,3 | 22,9 | 34,8 | 47,1 |
| 20 | – | – | 0,88 | 1,43 | 2,1 | 3,5 | 5,5 | 8,8 | 12,2 | 17,1 | 26,1 | 35,3 |
| 25 | – | – | – | 1,14 | 1,7 | 2,8 | 4,4 | 7,1 | 9,8 | 13,7 | 20,9 | 28,2 |
| 30 | – | – | – | – | 1,4 | 2,4 | 3,6 | 5,9 | 8,2 | 11,4 | 17,4 | 23,5 |
| 40 | – | – | – | – | – | 1,8 | 2,7 | 4,4 | 6,1 | 8,5 | 13 | 17,6 |
| 50 | – | – | – | – | – | – | 2,2 | 3,5 | 4,9 | 6,9 | 10,4 | 14,1 |
| 100 | – | – | – | – | – | – | – | 1,7 | 2,4 | 3,4 | 5,2 | 7,1 |
| 150 | – | – | – | – | – | – | – | – | – | 2,3 | 3,5 | 4,7 |
| 200 | – | – | – | – | – | – | – | – | – | – | 2,6 | 3,5 |
According to this table, our one and a half rack can only be 1 meter long. It will drop 2%, or 0.24V. We check using formula (4) - everything agrees.
If the voltage is higher (for example, 24 V DC), then the length can be correspondingly longer (2 times).
All of the above applies not only to constant voltage, but also to low voltage in general. And when choosing a cross-sectional area in such cases, one should be guided not only by the heating of the wire, but also by the voltage drop across it. For example, when powering halogen lamps through a step-down transformer.
Please comment on the article, how does theory coincide with practice?
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Calculation of voltage drop when feeding consumers with a loop
Calculating the voltage drop when powering consumers using radial circuits is quite simple. One section, one cable section, one length, one load current. We substitute this data into the formula and get the result.
When powering consumers via main circuits (loop), it is more difficult to calculate the voltage drop. In fact, you have to perform several voltage drop calculations for one line: you need to perform a voltage drop calculation for each section. Additional difficulties arise when the power consumption of electrical receivers powered by the main circuit changes. A change in the power of one electrical receiver is reflected in the entire chain.
How common is it in practice to supply power via main circuits and loops? There are many examples that can be given:
- In group networks, these are lighting networks and socket networks.
- In residential buildings, floor panels are powered using main circuits.
- In industrial and commercial buildings, main power supply circuits and panel loop power supply are also often used.
- The busbar is an example of supplying consumers via a trunk circuit.
- Power supply for outdoor road lighting poles.
Let's consider calculating the voltage drop using the example of outdoor lighting. Let's assume that you need to calculate the voltage drop for four outdoor lighting poles, sequentially powered from the ShchNO outdoor lighting panel.
The length of the sections from the shield to the pillar, between the pillars: L1, L2, L3, L4. Current flowing through sections: I1, I2, I3, I4. Voltage drop in sections: dU%1, dU%2, dU%3, dU%4.
Current consumed by the lamps on each pole, Ilamp.
The pillars are powered by a loop, respectively:
- I4=Ilamp
- I3=I4+Ilamp
- I2=I3+Ilamp
- I1=I2+Ilamp
The current consumed by the lamp is unknown, but the power of the lamp and its type are known (either from the catalog or according to clause 6.30 of SP 31-110-2003).
The current is determined by the formula:
Formula for calculating total phase current
Iph - total phase current P - active power
Uph - phase voltage
Cosφ - power factor
Nph - number of phases (Nph=1 for single-phase load, Nph=3 for single-phase load)
Let me remind you that the linear (phase-to-phase) voltage is √3 times greater than the phase voltage:
When calculating the voltage drop in a three-phase network, the line voltage drop is assumed, in single-phase networks - single-phase.
The voltage drop is calculated using the formulas:
Iph - total phase current flowing through the section R - section resistance
cosφ - power factor
The resistance of the section is calculated using the formula ρ - resistivity of the conductor (copper, aluminum) L - length of the section S - cross-section of the conductor
N - number of parallel conductors in the line
Typically, catalogs provide specific resistance values for various conductor cross-sections. If information about the specific resistance of conductors is available, the formulas for calculating the voltage drop take the form:
Formula for calculating voltage drop in a three-phase circuit
Formula for calculating voltage drop in a single-phase circuit
Substituting into the formula the corresponding values of currents, resistivities, length, number of parallel conductors and power factor, we calculate the magnitude of the voltage drop in the section.
Regulatory documents regulate the value of the relative voltage drop (as a percentage of the nominal value), which is calculated using the formula: U - rated network voltage.
The formula for calculating the relative voltage drop is the same for a three-phase and single-phase network. When calculating in a three-phase network, you need to substitute the three-phase drop and rated voltage, when calculating in a single-phase network - single-phase:
Formula for calculating the relative voltage drop in a three-phase network
Formula for calculating the relative voltage drop in a single-phase network
The theory is finished, let's look at how to implement this using DDECAD.
Let's take the following initial data:
- Lamp power 250W, cosφ=0.85.
- The distance between the pillars, from the shield to the first pillar is L1=L2=L3=L4=20m.
- The poles are powered by 3×10 copper cable.
- The branch from the power cable to the lamp is made with a 3×2.5 cable, L=6m.
For each column, we create a calculation table in the DDECAD program.
We fill in the data for the lamp in each calculation table: Connect the calculation table Column 3 to the calculation table Column 4, to Column 2 - Column 3, to Column 1 - Column 2, to SCHO - Column 1: Next, from the calculation table SCHO the drop value calculated by the program the voltage at the end of the first section (Column 1) is transferred to the green cell of the calculation table Column 1:
- In the calculation table Column 1, the cursor is placed on the green cell in the “∆U” column.
- Click "=".
- Switch to the SCHO calculation table.
- Place the cursor on the cell in the column “∆U∑”, located in the line Column 1.
- Press "Enter".
We get the calculated voltage drop at the end of the second section (Column 2) - 0.37% and the calculated voltage drop across the lamp - 0.27%.
We do the same for all other calculation tables and obtain the calculated values of the voltage drop in all sections. Since we linked the tables (using the program, connecting one table to another, and manually, transferring the voltage drop values), we got a linked system. When you make any changes, everything will be automatically recalculated.
ddecad.ru
Calculation of voltage drop along cable length
Power lines transport current from the switchgear to the final consumer along current-carrying conductors of varying lengths. The voltage will not be the same at the entry and exit points due to losses resulting from the large length of the conductor.
The voltage drop along the length of the cable occurs due to the passage of high current, causing an increase in the resistance of the conductor.
On lines of considerable length, losses will be higher than when current passes through short conductors of the same cross-section. To ensure that the required voltage is supplied to the final object, it is necessary to calculate the installation of lines taking into account losses in the current-carrying cable, starting from the length of the conductor.
Result of undervoltage
According to regulatory documents, losses on the line from the transformer to the most remote energy-loaded area for residential and public facilities should be no more than nine percent.
Losses of 5% are allowed to the main input, and 4% - from the input to the final consumer. For three-phase, three- or four-wire systems, the rating should be 400 V ± 10% under normal operating conditions.
Deviation of a parameter from the normalized value can have the following consequences:
- Incorrect operation of volatile installations, equipment, lighting devices.
- Failure of electrical appliances to operate when the input voltage is reduced, equipment failure.
- Reduced acceleration of the torque of electric motors at starting current, loss of energy taken into account, shutdown of motors when overheated.
- Uneven distribution of current load between consumers at the beginning of the line and at the remote end of a long wire.
- Lighting devices operate at half heat, resulting in underutilization of current power in the network and loss of electricity.
In operating mode, the most acceptable indicator of voltage loss in the cable is 5%. This is the optimal calculated value that can be accepted as acceptable for power grids, since in the energy industry currents of enormous power are transported over long distances.
Increased demands are placed on the characteristics of power lines. It is important to pay special attention to voltage losses not only on main networks, but also on secondary lines.
Causes of voltage drop
Every electromechanic knows that a cable consists of conductors - in practice, conductors with copper or aluminum cores wrapped in insulating material are used. The wire is placed in a sealed polymer shell - a dielectric housing.
Since the metal conductors are located too tightly in the cable and are additionally pressed by layers of insulation, when the power line is long, the metal cores begin to work on the principle of a capacitor, creating a charge with capacitive resistance.
The voltage drop occurs according to the following scheme:
- The conductor carrying the current overheats and creates capacitance as part of the reactance.
- Under the influence of transformations occurring on the windings of transformers, reactors, and other circuit elements, the power of electricity becomes inductive.
- As a result, the resistive resistance of the metal cores is converted into active resistance of each phase of the electrical circuit.
- The cable is connected to a current load with a total (complex) resistance along each current-carrying core.
- When operating a cable in a three-phase circuit, the three current lines in the three phases will be symmetrical, and the neutral core passes a current close to zero.
- The complex resistance of conductors leads to voltage losses in the cable when current flows with vector deviation due to the reactive component.
Graphically, the voltage drop diagram can be represented as follows: a straight horizontal line emerges from one point - the current vector. From the same point, the input voltage vector U1 and the output voltage vector U2 come out at an angle to the current at a smaller angle. Then the voltage drop along the line is equal to the geometric difference between the vectors U1 and U2.
Figure 1. Graphical representation of voltage drop
In the figure shown, the right triangle ABC represents the voltage drop and loss along a long cable line. Segment AB is the hypotenuse of a right triangle and at the same time the drop, legs AC and BC show the voltage drop taking into account active and reactance, and segment AD demonstrates the amount of losses.
It is quite difficult to make such calculations manually. The graph serves to visually represent the processes occurring in a long-distance electrical circuit when a current of a given load passes.
Calculation using formula
In practice, when installing trunk-type power lines and distributing cables to the end consumer with further distribution on site, copper or aluminum cable is used.
The resistivity for conductors is constant, for copper p = 0.0175 Ohm*mm2/m, for aluminum conductors p = 0.028 Ohm*mm2/m.
Knowing the resistance and current strength, it is easy to calculate the voltage using the formula U = RI and the formula R = p*l/S, where the following values are used:
- Wire resistivity - p.
- The length of the current-carrying cable is l.
- Conductor cross-sectional area - S.
- Load current in amperes - I.
- Conductor resistance - R.
- The voltage in the electrical circuit is U.
Using simple formulas on a simple example: it is planned to install several sockets in a detached extension of a private house. A copper conductor with a cross section of 1.5 square meters was selected for installation. mm, although for aluminum cable the essence of the calculations does not change.
Since the current passes back and forth through the wires, you need to take into account that the distance of the cable length will have to be doubled. If we assume that the sockets will be installed forty meters from the house, and the maximum power of the devices is 4 kW with a current of 16 A, then using the formula it is easy to calculate the voltage losses:
U = 0.0175*40*2/1.5*16
If we compare the obtained value with the nominal value for a single-phase line 220 V 50 Hz, it turns out that the voltage loss was: 220-14.93 = 205.07 V.
Such losses of 14.93 V are practically 6.8% of the input (nominal) voltage in the network. A value that is unacceptable for the power group of sockets and lighting fixtures, the losses will be noticeable: the sockets will pass current at less than full power, and the lighting fixtures will operate with less heat.
The power for heating the conductor will be P = UI = 14.93*16 = 238.9 W. This is the percentage of losses in theory without taking into account the voltage drop at the connection points of the wires and the contacts of the socket group.
Carrying out complex calculations
For a more detailed and reliable calculation of voltage losses on the line, it is necessary to take into account the reactive and active resistance, which together forms a complex resistance, and power.
To calculate the voltage drop in a cable, use the formula:
∆U = (P*r0+Q*x0)*L/ U nom
This formula contains the following values:
- P, Q - active, reactive power.
- r0, x0 - active, reactance.
- U nom - rated voltage.
To ensure optimal load on three-phase transmission lines, it is necessary to load them evenly. To do this, it is advisable to connect power electric motors to linear wires, and power to lighting devices - between the phases and the neutral line.
There are three load connection options:
- from the electrical panel to the end of the line;
- from the electrical panel with uniform distribution along the cable length;
- from the electrical panel to two combined lines with uniform load distribution.
An example of calculating voltage losses: the total power consumption of all volatile installations in a house or apartment is 3.5 kW - the average value for a small number of powerful electrical appliances. If all loads are active (all devices are connected to the network), cosφ = 1 (the angle between the current vector and the voltage vector). Using the formula I = P/(Ucosφ), the current strength is I = 3.5*1000/220 = 15.9 A.
Further calculations: if you use a copper cable with a cross section of 1.5 square meters. mm, resistivity 0.0175 Ohm*mm2, and the length of the two-core cable for wiring is 30 meters.
According to the formula, the voltage loss is:
∆U = I*R/U*100%, where the current is 15.9 A, the resistance is 2 (two wires)*0.0175*30/1.5 = 0.7 Ohm. Then ∆U = 15.9*0.7/220*100% = 5.06%.
The obtained value slightly exceeds the drop of five percent recommended by regulatory documents. In principle, you can leave the diagram for such a connection, but if the main values of the formula are affected by an unaccounted factor, the losses will exceed the permissible value.
What does this mean for the end consumer? Payment for used electricity supplied to the distribution panel at full capacity when actually consuming lower voltage electricity.
Using ready-made tables
How can a home craftsman or specialist simplify the calculation system when determining voltage losses along the cable length? You can use special tables given in highly specialized literature for power line engineers. The tables are calculated based on two main parameters - cable length of 1000 m and current value of 1 A.
As an example, a table is presented with ready-made calculations for single-phase and three-phase electrical power and lighting circuits made of copper and aluminum with different cross-sections from 1.5 to 70 square meters. mm when power is supplied to the electric motor.
Table 1. Determination of voltage loss along the cable length
| Nutrition | Lighting | Nutrition | Lighting | ||||
| Mode | Start | Mode | Start | ||||
| Copper | Aluminum | Cosine of phase angle = 0.8 | Cosine of phase angle = 0.35 | Cosine of phase angle = 1 | Cosine of phase angle = 0.8 | Cosine of phase angle = 0.35 | Cosine of phase angle = 1 |
| 1,5 | 24,0 | 10,6 | 30,0 | 20,0 | 9,4 | 25,0 | |
| 2,5 | 14,4 | 6,4 | 18,0 | 12,0 | 5,7 | 15,0 | |
| 4,0 | 9,1 | 4,1 | 11,2 | 8,0 | 3,6 | 9,5 | |
| 6,0 | 10,0 | 6,1 | 2,9 | 7,5 | 5,3 | 2,5 | 6,2 |
| 10,0 | 16,0 | 3,7 | 1,7 | 4,5 | 3,2 | 1,5 | 3,6 |
| 16,0 | 25,0 | 2,36 | 1,15 | 2,8 | 2,05 | 1,0 | 2,4 |
| 25,0 | 35,0 | 1,5 | 0,75 | 1,8 | 1,3 | 0,65 | 1,5 |
| 35,0 | 50,0 | 1,15 | 0,6 | 1,29 | 1,0 | 0,52 | 1,1 |
| 50,0 | 70,0 | 0,86 | 0,47 | 0,95 | 0,75 | 0,41 | 0,77 |
Tables are convenient to use for calculations when designing power lines. Calculation example: the motor operates with a rated current of 100 A, but at start-up a current of 500 A is required. During normal operation, cos ȹ is 0.8, and at start-up the value is 0.35. The electrical panel distributes a current of 1000 A. Voltage losses are calculated using the formula ∆U% = 100∆U/U nominal.
The engine is designed for high power, so it is rational to use a wire with a cross-section of 35 square meters for connection. mm, for a three-phase circuit in normal engine operation, the voltage loss is 1 volt over a wire length of 1 km. If the wire length is shorter (for example, 50 meters), the current is 100 A, then the voltage loss will reach:
∆U = 1 V*0.05 km*100A = 5 V
The losses at the switchboard when starting the engine are 10 V. The total drop is 5 + 10 = 15 V, which as a percentage of the nominal value is 100 * 15 * / 400 = 3.75%. The resulting number does not exceed the permissible value, so the installation of such a power line is quite realistic.
At the time of starting the engine, the current should be 500 A, and during operating mode - 100 A, the difference is 400 A, by which the current in the distribution board increases. 1000 + 400 = 1400 A. Table 1 indicates that when starting the engine, the losses along a cable length of 1 km are equal to 0.52 V, then
∆U at startup = 0.52*0.05*500 = 13 V
∆U shield = 10*1400/100 = 14 V
∆U total = 13+14 = 27 V, as a percentage ∆U = 27/400*100 = 6.75% - permissible value, does not exceed the maximum value of 8%. Taking into account all the parameters, the installation of the power line is acceptable.
Using the service calculator
Calculations, tables, graphs, diagrams - precise tools for calculating the voltage drop along the cable length. You can make your work easier if you perform the calculations using an online calculator. The advantages are obvious, but it is worth checking the data on several resources and starting from the average value obtained.
How it works:
- The online calculator is designed to quickly perform calculations based on initial data.
- You need to enter the following quantities into the calculator - current (alternating, direct), conductor (copper, aluminum), line length, cable cross-section.
- Be sure to enter parameters for the number of phases, power, network voltage, power factor, line operating temperature.
- After entering the initial data, the program determines the voltage drop along the cable line with maximum accuracy.
- An unreliable result can be obtained if the initial values are entered incorrectly.
You can use such a system to carry out preliminary calculations, since service calculators on various resources do not always show the same result: the result depends on the competent implementation of the program, taking into account many factors.
However, you can carry out calculations on three calculators, take the average value and build on it at the preliminary design stage.
How to cut losses
Obviously, the longer the cable on the line, the greater the resistance of the conductor when current passes and, accordingly, the higher the voltage loss.
There are several ways to reduce the percentage of losses that can be used either independently or in combination:
- Use a cable with a larger cross-section, carry out calculations in relation to a different conductor. An increase in the cross-sectional area of current-carrying conductors can be obtained by connecting two wires in parallel. The total cross-sectional area will increase, the load will be distributed evenly, and the voltage loss will be lower.
- Reduce the working length of the conductor. The method is effective, but it cannot always be used. The cable length can be reduced if there is a spare conductor length. At high-tech enterprises, it is quite realistic to consider the option of re-laying the cable if the costs of the labor-intensive process are much lower than the costs of installing a new line with a large cross-section of cores.
- Reduce the current power transmitted through long cables. To do this, you can disconnect several consumers from the line and connect them via a bypass circuit. This method is applicable on well-branched networks with backup highways. The lower the power transmitted through the cable, the less the conductor heats up, the resistance and voltage loss are reduced.
Attention! When the cable is operated at elevated temperatures, the conductor heats up and the voltage drop increases. Losses can be reduced by using additional thermal insulation or laying the cable along another route, where the temperature is significantly lower.
Calculation of voltage losses is one of the main tasks of the energy industry. If for the end consumer the voltage drop on the line and power losses are almost unnoticeable, then for large enterprises and organizations involved in supplying electricity to facilities, they are impressive. The voltage drop can be reduced if all calculations are performed correctly.
220.guru
How to calculate the voltage loss in a cable?
- Calculation of voltage loss for DC networks 12, 24, 36V.
- Calculation of voltage loss without taking into account inductive reactance 220/380V.
- Calculation of voltage loss taking into account inductive reactance 380V.
When designing networks, it is often necessary to calculate the voltage loss in the cable. Now I want to talk about the basic calculations of voltage loss in DC and AC networks, in single-phase and three-phase networks.
Let's turn to the regulatory documents and see what the permissible voltage deviation values are.
TKP 45-4.04-149-2009 (RB).
9.23 Deviations of voltage from the nominal voltage at the terminals of power electrical receivers and the most remote electric lighting lamps should not exceed ±5% in normal mode, and ±10% in after emergency mode at the highest design loads. In voltage networks
12–42 V (counting from a voltage source, for example a step-down transformer), voltage deviations are allowed to be accepted up to 10%.
Voltage deviation for electric motors in starting modes is allowed, but not more than 15%. In this case, stable operation of the starting equipment and engine starting must be ensured.
In normal operating mode, when power transformers are loaded into transformer substations not exceeding 70% of their rated power, the permissible (available) total voltage losses from the 0.4 kV busbars of the transformer substation to the most distant general lighting lamp in residential and public buildings, taking into account no-load losses transformers and voltage losses in them, reduced to secondary voltage, should, as a rule, not exceed 7.5%. At the same time, voltage losses in electrical installations inside buildings should not exceed 4% of the rated voltage, for stage lighting - 5%.
SP 31-110-2003 (RF). 7.23 Voltage deviations from the nominal voltage at the terminals of power electrical receivers and the most remote electric lighting lamps should not exceed ±5% in normal mode, and the maximum permissible in post-emergency mode at the highest design loads is ±10%. In networks with a voltage of 12-50 V (counting from a power source, for example a step-down transformer), voltage deviations are allowed to be accepted up to 10%.
For a number of electrical receivers (control devices, electric motors), a 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 busbars of the transformer substation to the most distant general lighting lamp in residential and public buildings should, as a rule, not exceed 7.5%.
The range of voltage changes at the terminals of electrical receivers when starting an electric motor should not exceed the values established by GOST 13109.
5.3.2 The maximum permissible value of the sum of the steady-state voltage deviation dUy and the range of voltage changes at points of connection to electrical networks with a voltage of 0.38 kV is equal to 10% of the rated voltage.
Voltage loss depends on the cable material (copper, aluminum), cross-section, line length, power (current) and voltage.
To calculate the voltage loss, I made 3 programs in Excel based on the book by F.F. Karpov “How to choose the cross-section of wires and cables.”
1 For DC networks, inductive reactance is not taken into account. The voltage loss can be calculated using the following formulas (for a two-wire line):
Using these formulas, I calculate the voltage loss of electric drives for opening windows (24V), as well as the lighting network (220V).
Appearance of the program for calculating voltage loss 12, 24, 36, 42V
2 For three-phase networks, where the cosine is 1, inductive reactance is also not taken into account. This method can also be used for lighting networks because... their cos is close to 1, the error we get is not significant. Formula for calculating voltage loss (380V):
Appearance of the program for calculating voltage loss 220/380V
3 Calculation of voltage loss taking into account inductive reactance is used in other cases, in particular in networks. Formula for calculating voltage loss taking into account inductive reactance:
Appearance of the program for calculating voltage loss 380V, 6kV, 10kV
To get the program, go to the MY PROGRAMS page.
I look forward to your feedback and suggestions :)
Long cable lines are characterized by significant resistance, which makes adjustments to the operation of the network. Depending on the brand of cable and other parameters, the resistance value will also differ. And the amount of voltage on the cable line is directly proportional to this resistance.
Using an online calculator, calculating voltage losses in a cable comes down to the following steps:
- Indicate the cable length in meters and the material of the current-carrying conductors in the appropriate boxes;
- Conductor cross-section in mm²;
- The amount of electricity consumed in amperes or watts (place the indicator next to the power or current, depending on what parameter you know and what value you will indicate);
- Enter the voltage value in the network;
- Enter the power factor cosφ;
- Specify the cable temperature;
After you have entered the above data into the fields of the calculator, click the “calculate” button and in the corresponding columns you will receive the calculation result - the amount of voltage loss in the cable ΔU in%, the resistance of the wire itself R pr in Ohm, reactive power Q pr in VAR and voltage at load U n.
To calculate these values, the entire system, including cable and load, is replaced with an equivalent one, which can be represented as follows:
As you can see in the figure, depending on the type of power supply to the load (single-phase or three-phase), the resistance of the cable line will have a series or parallel connection with respect to the load. Calculation in the calculator is carried out using the following formulas:
- ΔU – voltage loss;
- U L – linear voltage;
- U Ф – phase voltage;
- I – current flowing in the line;
- Z K – cable line impedance;
- R K – active resistance of the cable line;
- X K – cable line reactance.
Of these, U L, U F, I, are specified at the data entry stage. To determine the total resistance Z K, the arithmetic addition of its active R K and reactive X K components is performed. Active and reactive resistance is determined by the formulas:
R K = (ρ * l) / S
R K – active resistance of the cable line, where
ρ is the resistivity for the corresponding metal (copper or aluminum), but the value of the resistivity of the material is not constant and can vary depending on the temperature, which is why, to bring it to real conditions, recalculation is performed in relation to temperature:
ρ t = ρ 20 *
- a is the coefficient of temperature change in the resistivity of the material.
- ρ 20 – specific resistance of the material at a temperature of +20ºС.
- t is the actual temperature of the conductor at a given time.
- l – cable line length (if the load is single-phase and the cable has two cores, then both of them are connected in series and the length must be multiplied by 2)
- S – cross-sectional area of the conductor.
Reactive power is determined by the following formula: Q = S*sin φ, where
Where S is the apparent power, which can be defined as the product of the current in the circuit and the input voltage of the source or as the ratio of active power to power factor.
To calculate the voltage per load, the following calculations are made: U H = U - ΔU, where
- Where U N is the magnitude of the voltage applied to the load;
- U – voltage at the input to the cable line
- ΔU – voltage drop in the cable line.
Main switchboard 2.2. Indications of phase voltages after the first section of the cable line
Backup power supply parameters:
- Maximum power of diesel power plant – 600 kW,
- Cable line – 3 cables AVBbShv 4x240, connected in parallel,
- Cable line length – 250 m.
Based on these parameters, we can clearly conclude that the capacity of the diesel power plant and the backup cable line, taking into account the voltage drop, will be enough for no more than half of the maximum load requirements, which is completely unacceptable.
Therefore, it makes no sense to monitor the quality of food through the diesel power station.
Download file
In conclusion - as promised, a good book on calculating voltage loss and voltage loss in a cable. It will be very interesting to everyone who is interested in this article. Nowadays such books are no longer written.
/ Brochure from the Electrician's Library. Provides instructions and calculations necessary for selecting cross-sections of wires and cables up to 1000 V. Useful for those who are interested in primary sources., zip, 1.57 MB, downloaded: 385 times./
