Radiation type low-voltage power distribution network loss calculation method based on equivalent resistance method

Abstract

The invention provides a radiation type low-voltage power distribution network loss calculation method based on an equivalent resistance method, relates to the field of radiation type low-voltage power distribution network loss calculation, and solves the technical problems that a reasonable resistance equivalent model cannot be established for a radiation type power distribution network and the line loss of the power distribution network cannot be accurately calculated in the prior art. The method comprises the following steps: dividing a power distribution network into n sections, calculating primary line equivalent resistance according to the sum of resistance load current of each section and the sum of line power distribution loss, and performing temperature correction to obtain line equivalent resistance; calculating the equivalent resistance of the primary transformer according to the sum of the load current and the sum of the load loss power of each section of transformer, and correcting according to the three-phase unbalance degree to obtain the equivalent resistance of the transformer; and calculating the total electric energy loss of the power distribution network in the operation time of the power distribution network based on the line equivalent resistance and the transformer equivalent resistance. The method is used in the process of calculating the network loss of the radiation type low-voltage power distribution network.

Description

Technical Field

[0001] This application relates to the field of line loss calculation for radial distribution networks, and in particular to a method for calculating network losses in radial low-voltage distribution networks based on the equivalent resistance method. Background Technology

[0002] Line losses in distribution networks are unavoidable energy losses in power system operation. They mainly originate from resistance losses due to current flowing through lines, insulation leakage, and corona discharge. These losses are widespread in the transmission of electricity to users and are an inherent characteristic of distribution network operation. Accurate line loss calculation is the foundation for developing loss reduction measures. By quantifying the sources and distribution of losses, targeted technical means can be adopted to reduce energy consumption, helping to achieve energy conservation and emission reduction goals and promoting the sustainable development of the power industry.

[0003] Currently, there are two main methods for calculating theoretical line losses in radial distribution networks: one is the root-mean-square (RMS) current method, which assumes that the energy loss caused by the RMS current through the conductor is equal to the energy loss caused by all charges in the same time period. The other is the average current method, also known as the shape factor method, which calculates power loss using the equivalent relationship between the RMS current and the average current. The RMS current method assumes an equivalent total charge loss based on the RMS current, neglecting the details of load dynamic fluctuations; the average current method simplifies the relationship between the RMS current and the average current using a shape factor, but the shape factor itself is difficult to obtain accurately and often requires further simplification. The simplification assumptions in these studies deviate significantly from actual scenarios and differ somewhat from the actual operating conditions of the distribution network. Therefore, proposing appropriate simplification conditions and establishing a reasonable resistance equivalent model for radial distribution networks to accurately calculate line losses and optimize grid management has become an urgent technical problem to be solved. Summary of the Invention

[0004] This application provides a method for calculating the network loss of radial low-voltage distribution networks based on the equivalent resistance method, which solves the technical problem that existing technologies cannot establish a reasonable equivalent resistance model for radial distribution networks and cannot accurately calculate the line losses of distribution networks.

[0005] To achieve the above objectives, this application adopts the following technical solution: A method for calculating the losses in a radial low-voltage distribution network based on the equivalent resistance method is provided, including: The power distribution network is divided into n segments, and the line parameters and transformer data of each segment are collected; the line parameters include resistive load current and line distribution loss; the transformer data includes load current and load loss power. After calculating the elementary equivalent resistance of the line by summing the resistive load current of each segment and summing the line distribution loss, the equivalent resistance of the line is obtained by temperature correction. After calculating the elementary transformer equivalent resistance by summing the load current and summing the load loss power of each transformer segment, the transformer equivalent resistance is obtained by correcting it according to the three-phase unbalance. Calculate the total power loss of the distribution network during its operating time based on the equivalent resistance of the lines and transformers.

[0006] Based on the above technical solution, the method for calculating losses in radial low-voltage distribution networks based on the equivalent resistance method provided in this application addresses the characteristics of radial low-voltage distribution networks by employing a dual-correction logic: temperature-corrected line equivalent resistance and three-phase unbalanced correction of transformer equivalent resistance. This overcomes the shortcomings of traditional loss calculations that neglect the influence of ambient temperature on line resistance and the amplification effect of three-phase load imbalance on transformer losses. This makes the calculated losses of lines and transformers closer to actual operating conditions, significantly reducing the calculation error of total power loss. The required parameters, such as line resistance, transformer rated capacity, and average load factor, are all readily available nameplate data and smart meter metering data from distribution network operation and maintenance. No additional complex monitoring equipment is required, reducing the implementation cost and technical threshold of the method, and making it more suitable for the actual application conditions of grassroots distribution networks.

[0007] By employing the equivalent resistance method, the losses of dispersed multi-segment lines and multiple transformers are represented as a concentrated loss model of the total equivalent resistance of the lines and transformers. Combined with operating time, the total power loss can be quickly derived, avoiding the tedious calculation of each node and branch in the distribution network, significantly improving the efficiency of loss calculation. This method is suitable for batch calculations of large-scale radial low-voltage distribution networks. It also covers variable line losses, transformer load losses, and transformer no-load losses, ultimately outputting the total power loss over the operating time. This comprehensively covers the core loss types of distribution networks, and the results can directly serve the practical needs of distribution network line loss statistics, energy-saving retrofit assessments, and operation optimization, possessing strong engineering guidance value.

[0008] In one possible implementation, the equivalent resistance of the line can be obtained in the following ways: Based on temperature correction factor The equivalent resistance of the elementary circuit is corrected to obtain the equivalent resistance of the circuit; the formula for calculating the equivalent resistance of the circuit is:

[0009] in, The equivalent resistance of the line. For elementary circuits, the equivalent resistance is... This represents the actual ambient temperature.

[0010] By introducing a temperature correction coefficient, the shortcomings of traditional methods in ignoring line resistance shifts caused by changes in ambient temperature are compensated for, making the equivalent line resistance more closely match the actual operating environment and further improving the accuracy of loss calculation. The temperature correction logic is adapted to the actual characteristics of low-voltage distribution networks laid outdoors and subject to ambient temperature fluctuations, making it particularly suitable for loss calculations in distribution networks in different seasons and regions, thus enhancing the universality of the technical solution.

[0011] In one possible implementation, the formula for calculating the equivalent resistance of an elementary transformer is:

[0012] in, This is the equivalent resistance of an elementary transformer. This represents the total power loss due to the load. This represents the total load current of the transformer.

[0013] Using formula The method for calculating the equivalent resistance of elementary transformers is simple in structure and easy to obtain parameters, making it convenient for rapid application in practical engineering. At the same time, it provides a reliable base value for subsequent three-phase imbalance correction.

[0014] In one possible implementation, the equivalent resistance of the transformer is obtained by: calculating the equivalent resistance of the transformer using the elementary equivalent resistance and the total three-phase unbalance; the formula for calculating the equivalent resistance of the transformer is:

[0015] in, The equivalent resistance of the transformer. This is the equivalent resistance of an elementary transformer. This represents the total three-phase imbalance.

[0016] The method for obtaining the total three-phase imbalance is as follows: obtain the transformer current of the phase with the largest phase imbalance. and the average value of the three-phase current of the transformer The total three-phase imbalance is calculated; the total three-phase imbalance The formula for calculation is: .

[0017] A three-phase unbalance correction coefficient is introduced to compensate for the shortcomings of traditional methods that ignore the amplification effect of unbalanced operation on transformer losses, making the equivalent resistance of the transformer more closely reflect actual operating conditions and reducing errors in load loss calculation. The derivation is based on the inherent parameters on the transformer's nameplate, such as rated capacity and short-circuit voltage, eliminating the need for complex real-time monitoring data. Furthermore, it is adapted to the unbalanced operating characteristics of radial distribution networks, enhancing the relevance of the technical solution.

[0018] In one possible implementation, the formula for calculating the total power loss of the distribution network is:

[0019] in, The total power loss of the distribution network during the operating time period t. Let J be the no-load loss of the j-th distribution transformer in the distribution network. This is the equivalent current for distribution network losses.

[0020] The no-load loss of a transformer is the active power absorbed by the transformer from the power source when the rated voltage at the rated frequency is applied to the primary side of the transformer and the secondary side is open.

[0021] The no-load loss of a transformer is directly summed from the no-load loss nameplate parameters of a single transformer, eliminating the need for complex operating condition corrections. This solves the cumbersome problem of relying on real-time monitoring data in traditional methods, significantly improving calculation efficiency. By integrating the total variable line loss, total transformer no-load loss, and transformer load loss, the total power loss during the operating time is ultimately output without additional secondary calculations, directly serving practical needs such as line loss statistics and energy-saving benefit assessments. Clearly distinguishing the proportion of different types of losses helps maintenance personnel quickly locate high-loss-prone areas, providing clear guidance for developing targeted loss reduction measures.

[0022] In one possible implementation, the formula for calculating the equivalent resistance of elementary circuits is:

[0023] in, For elementary circuits, the equivalent resistance is... This represents the total power distribution loss of the line. This represents the sum of the currents from the resistive load.

[0024] pass The formula for calculating the equivalent resistance of elementary circuits is simple and directly related to the actual loss data of the circuit, which facilitates the rapid deduction of the equivalent resistance based on measured data, thus improving the operability and reliability of the calculation.

[0025] In one possible implementation, the method for obtaining the sum of resistive load currents includes: calculating the resistive load current of each segment using the line parameters and transformer data of each segment; summing the resistive load currents of all segments to obtain the sum of resistive load currents; the formula for calculating the sum of resistive load currents is:

[0026] in, This is the sum of the currents from the resistive load. For the first i The resistive load current of each segment.

[0027] The method for obtaining the resistive load current of each segment is as follows: In n segments of the distribution network, the nth segment... i The resistance corresponding to each segment If j distribution transformers are subsequently connected, the formula for calculating the resistive load current flowing through the aforementioned section is:

[0028] in, For the first i The resistive load current of each segment, Let j be the rated capacity of the j-th transformer. Let j be the average load factor of the j-th transformer. For the first i The number of transformers in each section.

[0029] By calculating and summing the resistive load current segment by segment, the differences in load distribution in each segment of the distribution network are fully considered, making the calculation of equivalent resistance more consistent with the actual network structure and improving the structural realism and calculation accuracy of the model.

[0030] In one possible implementation, the formula for calculating the total line distribution loss is as follows: Calculate the line distribution loss of each segment using the line parameters and transformer data; sum the line distribution losses of all segments to obtain the total line distribution loss; the total line distribution loss... The formula for calculation is:

[0031] The power distribution losses of each section of the line The formula for calculation is: .

[0032] By accurately matching the generation principle of variable losses in power lines through loss formula calculations, the disconnect between loss calculations and electrical principles is avoided. The total variable losses calculated over the operating time can be directly used for practical work such as evaluating energy-saving retrofits of power lines and locating the causes of losses, providing a quantitative basis for formulating loss reduction measures for distribution networks.

[0033] In one possible implementation, the method for obtaining the total load current of the transformers includes: calculating the load current of the j-th transformer based on the transformer segment data; summing the load currents of all transformers to obtain the total load current of the transformers; the formula for calculating the total load current of the transformers is:

[0034] in, This represents the total load current of the transformer. Let be the load current of the j-th transformer.

[0035] The formula for calculating the load current of the j-th transformer is:

[0036] in, Let J be the load current of the j-th transformer. Let j be the rated capacity of the j-th transformer. Let U be the average load factor of the j-th transformer. N This is the system's rated voltage.

[0037] By summing the load currents of each transformer to obtain the total current, and combining the rated capacity and load factor of the transformers, a reasonable estimate of the transformer-side current is achieved, providing accurate input data for the calculation of the equivalent resistance of the transformers.

[0038] In one possible implementation, the formula for calculating the total load loss power is:

[0039] in, This represents the total power loss due to the load. For steady-state short-circuit loss, Let be the average load factor of the j-th transformer.

[0040] The steady-state short-circuit loss of a transformer refers to the active power consumed by the transformer under steady-state conditions where one winding is short-circuited and the other winding is subjected to a rated frequency voltage, causing the short-circuit winding current to reach its rated value. This formula for the sum of load loss power fully considers the square relationship of the load rates of each transformer, conforms to the actual physical laws of transformer load loss, and improves the accuracy and engineering practicality of transformer loss calculations.

[0041] This application provides a method and apparatus for calculating network losses in radial low-voltage distribution networks based on the equivalent resistance method. It quantifies the equivalent resistance of radial distribution networks by constructing equivalent resistance values ​​for lines and transformers, considering both line losses and transformer losses. This allows for more accurate calculation of total network losses, accurately reflecting the operational efficiency of the distribution network, providing data support for evaluating the rationality of power grid planning, and promoting refined power grid management. By considering the resistance-temperature characteristics of low-voltage lines, a temperature correction coefficient is introduced to correct the equivalent resistance of the lines. Furthermore, a three-phase imbalance degree is proposed to correct the equivalent resistance of transformers, addressing the common three-phase load imbalance problem in radial low-voltage distribution networks. These improvements enhance the calculation accuracy of line resistance and transformer resistance, thereby increasing the reliability of the calculation results.

[0042] It should be understood that the descriptions of technical features, technical solutions, beneficial effects, or similar language in this application do not imply that all features and advantages can be achieved in any single embodiment. Rather, it is understood that the description of a feature or beneficial effect means that a specific technical feature, technical solution, or beneficial effect is included in at least one embodiment. Therefore, the descriptions of technical features, technical solutions, or beneficial effects in this specification do not necessarily refer to the same embodiment. Furthermore, the technical features, technical solutions, and beneficial effects described in this embodiment can be combined in any suitable manner. Those skilled in the art will understand that embodiments can be implemented without one or more specific technical features, technical solutions, or beneficial effects of a particular embodiment. In other embodiments, additional technical features and beneficial effects may be identified in specific embodiments that do not embody all embodiments. Attached Figure Description

[0043] Figure 1 A flowchart illustrating a method for calculating losses in a radial low-voltage distribution network based on the equivalent resistance method, provided in this application embodiment; Figure 2 A flowchart illustrating another method for calculating losses in a radial low-voltage distribution network based on the equivalent resistance method, provided for an embodiment of this application; Figure 3 A flowchart illustrating another method for calculating losses in a radial low-voltage distribution network based on the equivalent resistance method, provided for an embodiment of this application; Detailed Implementation

[0044] In the description of this application, unless otherwise stated, " / " means "or," for example, A / B can mean A or B. The "and / or" in this document is merely a description of the relationship between related objects, indicating that three relationships can exist. For example, A and / or B can represent: A alone, A and B simultaneously, and B alone. Furthermore, "at least one" means one or more, and "multiple" means two or more. The terms "first," "second," etc., do not limit the quantity or order of execution, and "first," "second," etc., do not necessarily imply differences.

[0045] It should be noted that, in this application, the terms "exemplary" or "for example" are used to indicate that something is being described as an example, illustration, or illustration. Any embodiment or design described as "exemplary" or "for example" in this application should not be construed as being more preferred or advantageous than other embodiments or design solutions. Specifically, the use of terms such as "exemplary" or "for example" is intended to present the relevant concepts in a concrete manner.

[0046] To address the technical problem in existing technologies where suitable simplification conditions are lacking, making it impossible to establish a reasonable equivalent resistance model for radial distribution networks and accurately calculate line losses, this application provides a method for calculating network losses in radial low-voltage distribution networks based on the equivalent resistance method. This method includes: dividing the distribution network into n segments and collecting line parameters and transformer data for each segment; the line parameters include resistive load current and line distribution losses; the transformer data includes load current and load loss power; calculating the initial equivalent line resistance by summing the resistive load current and summing the line distribution losses of each segment, and then performing temperature correction to obtain the equivalent line resistance; calculating the initial equivalent transformer resistance by summing the load current and summing the load loss power of each transformer segment, and then correcting based on three-phase imbalance to obtain the equivalent transformer resistance.

[0047] like Figure 1 As shown in the embodiments of this application, the method for calculating the network loss of a radial low-voltage distribution network based on the equivalent resistance method includes: S101. Divide the distribution network into n segments and collect the line parameters and transformer data of each segment.

[0048] The line parameters include resistive load current and line distribution loss; transformer data includes load current and load loss power; the distribution network refers to the lines that transmit power from step-down substations to distribution transformers or from distribution substations to power-consuming units.

[0049] In some implementations, a trunk-and-radial topology of the radial distribution network is used, and the division must follow the radial path from the power source to the load, thus breaking it down segment by segment. The trunk-and-radial topology uses substations as the root, lines as the trunk or branches, and transformers as leaf loads. First, starting from a step-down substation or distribution substation, the main trunk line of the radial network is identified. This trunk line is the main path carrying the core power supply to the entire area, typically the line with the largest conductor cross-section and the longest power supply distance. Second, the trunk line is divided into basic segments through T-junctions, bus branch points, and points where conductor parameters change. Each basic segment is a continuous line segment with consistent parameters. Then, the number and specific affiliation of the distribution transformers connected downstream of each segment need to be clearly defined to ensure a unique correspondence between segments and transformers. Finally, the rationality of the distribution network division needs to be verified to ensure it is calculable and non-overlapping, avoiding errors in subsequent loss calculations.

[0050] It should be noted that the essence of distribution network segmentation is to decompose a complex radial network into calculable independent units. All criteria serve the accuracy and operability of subsequent equivalent resistance derivation and loss calculation. Furthermore, the segmentation of distribution network sections must revolve around three core criteria: structural characteristics, load correlation, and parameter consistency. The segmentation method must conform to the trunk-radial topological characteristics of radial distribution networks.

[0051] S102. After calculating the elementary equivalent resistance of the line by the sum of the resistive load current of each section and the sum of the line distribution loss, the equivalent resistance of the line is obtained by temperature correction.

[0052] Among these losses, the active power loss caused by the resistance of the conductors when current flows through the distribution line can be equivalent to the current passing through the equivalent resistance of an elementary line. Loss; The equivalent resistance of the circuit at a rated temperature of 20℃; resistance temperature correction factor. The actual ambient temperature (T) is obtained by checking the wire type; the actual ambient temperature (T) is obtained from the temperature sensor in the transformer area.

[0053] In some implementations, the formula for calculating the equivalent resistance of elementary circuits is:

[0054] in, For elementary circuits, the equivalent resistance is... This represents the total power distribution loss of the line. This represents the sum of the currents from the resistive load.

[0055] It should be noted that the calculation of the equivalent resistance of elementary circuits must ensure... and Corresponding to the same runtime segment, and The loss must include all line segments; if the current difference between different segments is large, the loss must be calculated segment by segment and then summed. Otherwise, incomplete loss statistics will lead to... If the value is too small, it will affect the accuracy of the subsequent calculation of total loss.

[0056] It should also be noted that if the average load factor of all distribution transformers in the network is the same, k i =k1=k2=…=k m Then the calculation formula can be used. Simplify to .

[0057] For example, suppose that during the operating period of a certain distribution network, the total line distribution loss is... =2880W, corresponding to the total resistive load current during the same period. =80A, substituting into the formula, we can obtain the equivalent resistance of the elementary circuit. =2880 / (3× =2880 / 19200 = 0.15Ω.

[0058] In some implementations, the equivalent resistance of the line is obtained by means of: based on a temperature correction factor. The equivalent resistance of the elementary circuit is corrected to obtain the equivalent resistance of the circuit; the formula for calculating the equivalent resistance of the circuit is: ;in, The equivalent resistance of the line. For elementary circuits, the equivalent resistance is... This represents the actual ambient temperature.

[0059] It should be noted that the resistance of low-voltage lines is significantly affected by ambient temperature, with the resistance increasing by approximately 4% for every 10°C increase in temperature. Therefore, introducing a temperature correction factor can improve the accuracy of line resistance calculations.

[0060] It should also be noted that the temperature correction factor The coefficient must be strictly matched with the material of the conductor; coefficients for different materials cannot be mixed. The actual ambient temperature T should be the average ambient temperature corresponding to the period of loss calculation to avoid correction deviations caused by extreme temperatures. The 20 in the formula is the industry-standard rated reference temperature for conductors. If the design documents specify other reference temperatures, this value must be replaced accordingly.

[0061] For example, setting the equivalent resistance of elementary circuits. =0.15Ω, the circuit uses copper wire. 0.004 / The average ambient temperature T during actual operation was 35°C. Then the equivalent resistance of the line =0.15 [1+0.004 (35 20)]=0.15 1.06 = 0.159Ω.

[0062] S103. After calculating the elementary transformer equivalent resistance by summing the load current and summing the load loss power of each transformer segment, the transformer equivalent resistance is obtained by correcting it according to the three-phase unbalance.

[0063] Among them, during the operation of a distribution transformer, the active power loss caused by the load current in the winding resistance can be equivalent to the current passing through the transformer's equivalent resistance. The losses. The current in each phase is measured by the system's measuring device; steady-state short-circuit losses. The parameters are given on the nameplate of the distribution transformer.

[0064] In some implementations, the method for obtaining the total load current of the transformers includes: calculating the load current of the j-th transformer based on the transformer segment data; summing the load currents of all transformers to obtain the total load current of the transformers; the formula for calculating the total load current of the transformers is:

[0065] in, This represents the total load current of the transformer. Let be the load current of the j-th transformer.

[0066] The formula for calculating the load current of the j-th transformer is:

[0067] in, Let J be the load current of the j-th transformer. Let j be the rated capacity of the j-th transformer. Let U be the average load factor of the j-th transformer. N This is the system's rated voltage.

[0068] For example, assume the rated capacity of the j-th transformer is... For a 100kVA transformer, the average load factor of the j-th transformer is... Given a value of 0.6 and a system rated voltage UN of 0.4kV, substituting these values ​​into the formula yields:

[0069] It enables precise calculation of the load current of a single transformer, avoiding the current distortion of a single device caused by the traditional method of estimating the total current based on the total capacity. This provides accurate equipment-level current parameters for the subsequent derivation of transformer load loss and equivalent resistance.

[0070] In some implementations, the formula for calculating the total load loss power is:

[0071] in, This represents the total power loss due to the load. This refers to the steady-state short-circuit loss of the transformer. Let be the average load factor of the j-th transformer.

[0072] It should be pointed out that, It is based on active power data during the same operating period. and reactive power data Derived, that is To avoid load factor deviations caused by mixing data from different time periods; The steady-state short-circuit loss indicated on the transformer nameplate must be used. If the distribution network contains multiple transformer models, the nameplate parameters of each transformer must be taken individually, and a single value cannot be used uniformly.

[0073] For example, suppose a power distribution network contains two transformers: the first transformer... =2.5kW =0.6, the second unit =3.8kW =0.55, then the total load loss power is... =

[0074] .

[0075] In some implementations, the formula for calculating the equivalent resistance of an elementary transformer is:

[0076] in, This is the equivalent resistance of an elementary transformer. This represents the total power loss due to the load. This represents the total load current of the transformer.

[0077] It should be noted that the total transformer load current Total power loss with load The data must be obtained during the same operating period to ensure the matching of current and losses. The 3 in the formula is a loss correction factor for a three-phase system, corresponding to 3 times the single-phase loss. This only applies to three-phase distribution transformers; for single-phase transformers, the factor should be adjusted to 1. The equivalent resistance of a primary transformer is an ideal value that ignores three-phase imbalance and needs to be corrected for imbalance to reflect actual operating losses.

[0078] It should also be noted that if the average load factor of all distribution transformers in the network is the same, then according to the definition of the equivalent resistance of distribution transformers, combined with these two formulas: , The formula for calculating the equivalent resistance of an elementary transformer can be directly derived: .

[0079] For example, if the total transformer load current during the same period is 100A, total load loss power If the value is 2070W, then the equivalent resistance of the elementary transformer is... .

[0080] In some implementations, the equivalent resistance of the transformer is obtained by: calculating the equivalent resistance of the transformer using the elementary equivalent resistance and the total three-phase unbalance; the formula for calculating the equivalent resistance of the transformer is:

[0081] in, The equivalent resistance of the transformer. This is the equivalent resistance of an elementary transformer. This represents the total three-phase imbalance.

[0082] The method for obtaining the total three-phase imbalance is as follows: obtain the transformer current of the phase with the largest phase imbalance. and the average value of the three-phase current of the transformer The total three-phase imbalance is calculated; the total three-phase imbalance The formula for calculation is: .

[0083] It should be noted that the maximum phase current of the transformer and the average value of the three-phase current of the transformer The effective values ​​of the three-phase currents at the same moment should be used to avoid distortion of the unbalance caused by data across different time periods; The industry-permissible threshold is typically 20%. Exceeding this threshold will significantly increase the additional losses of the transformer. In the above formula, 0.02 is an empirical correction coefficient for low-voltage distribution networks, which corresponds to an increase of approximately 2% in transformer losses for every 1% increase in imbalance. If the distribution network has a specific imbalance correction standard, this coefficient can be replaced.

[0084] It should also be noted that low-voltage distribution transformers are mostly three-phase four-limb type. Three-phase load imbalance will lead to additional losses, which account for about 5%-15% of the total losses. Therefore, considering the three-phase load imbalance when calculating the equivalent resistance of the transformer can improve its calculation accuracy.

[0085] For example, suppose the three-phase current of a transformer is: Elementary transformer equivalent resistance = ,but =110A、 =(95+110+95) / 3=110A, total three-phase imbalance =(110 100) / 100 100% = 10%, therefore, .

[0086] S104. Calculate the total power loss of the distribution network during its operating time based on the equivalent resistance of the lines and the equivalent resistance of the transformers.

[0087] Among them, the no-load loss of the j-th distribution transformer in the distribution network The parameters are given on the nameplate of the distribution transformer.

[0088] In some implementations, the formula for calculating the total power loss of the distribution network is:

[0089] in, The total power loss of the distribution network during the operating time period t. Let J be the no-load loss of the j-th distribution transformer in the distribution network. This is the equivalent current for distribution network losses.

[0090] It should be noted that the equivalent current of distribution network losses The formula for calculation is: ,in For a certain time period t, the active energy meter reading W and the reactive energy meter reading W' are... q Calculated average current The product of the current shape factor h. The current shape factor h is used for correction. and The waveform deviation between W and W's needs to be determined based on the actual load characteristics. Because residential loads account for a high proportion in radial low-voltage distribution networks, the current waveform fluctuates significantly with electricity consumption behavior, such as large differences in peak current values ​​between morning and evening peak hours. Therefore, a current shape factor h is needed to adjust the waveform deviation. q It can be obtained through smart meters or metering devices at the incoming line of the power distribution network.

[0091] It should be noted that the equivalent current of distribution network losses Equivalent resistance of the line and transformer equivalent resistance Obtained within the same working period. The no-load loss of the j-th distribution transformer in the distribution network. The no-load loss parameters must be obtained from the nameplates of the distribution transformers, and must cover all transformers in the entire network. If The calculation result is in W, which needs to be converted to kW before being compared with the result in kW. Add them together, and take the running time t as hours, to get the final total power loss. The unit is kWh.

[0092] It should also be noted that in radial low-voltage distribution networks, transformers are often under light load or no-load conditions, and the proportion of no-load losses may be significantly increased. Accurately accounting for no-load losses is necessary to more comprehensively reflect the actual energy loss of the distribution network, providing a quantitative basis for transformer selection and operation scheduling, and helping to reduce overall line losses.

[0093] For example, if the equivalent resistance of the line is known 0.159Ω, equivalent resistance of a transformer The equivalent current of distribution network loss is 17.39Ω. The current rating is 80A, and the no-load losses of the two transformers are respectively... For 0.3kW, Given a power of 0.4 kW and an operating time of t of 24 hours, substituting these values ​​into the formula yields the following result: 336.94kW; .

[0094] Therefore, the total power loss is: .

[0095] Based on the above technical solution, the method for calculating losses in radial low-voltage distribution networks based on the equivalent resistance method provided in this application clearly distinguishes distribution network losses into line losses and transformer losses, and constructs equivalent resistance models for each, thus achieving structured modeling of the loss sources. This method overcomes the shortcomings of traditional root-mean-square current methods or average current methods, which ignore network structure differences due to overall equivalence, and can more realistically reflect the actual loss distribution of radial distribution networks, significantly improving the overall accuracy of line loss calculation.

[0096] In the calculation of the equivalent resistance of power lines, a temperature correction coefficient is introduced to effectively reflect the actual impact of ambient temperature on conductor resistance. This correction enables the model to adapt to temperature changes in different seasons and climate regions, making it particularly suitable for areas with large diurnal temperature differences or significant seasonal temperature fluctuations, thereby improving the environmental robustness and practicality of the calculation results. For the common three-phase load imbalance problem in low-voltage distribution networks, this invention proposes a transformer equivalent resistance correction method based on the total three-phase imbalance. This method fully considers the additional losses caused by unbalanced loads, making the transformer loss calculation more closely reflect actual operating conditions, and is particularly suitable for application scenarios with large load fluctuations and significant three-phase imbalances, such as residential and commercial areas.

[0097] By calculating line losses with high precision, this invention can provide reliable data support for power grid planning, operation scheduling, energy efficiency assessment, and loss reduction upgrades. Combined with the inclusion of no-load losses, it can comprehensively assess the actual energy consumption level of the distribution network, providing a quantitative basis for formulating energy-saving measures such as transformer selection, load adjustment, and reactive power compensation, thus helping the power system achieve refined and intelligent management.

[0098] In one possible implementation of the embodiments of this application, combined with Figure 1 ,like Figure 2 As shown, the above S102 can be implemented by the following S201 and S202, which are explained in detail below: S201. Calculate the resistive load current of each section using the line parameters and transformer data of each section.

[0099] Among them, the rated capacity of the j-th transformer The average load factor of the j-th transformer is given directly from the transformer nameplate parameters. Provided by real-time operational monitoring data.

[0100] In some implementations, the resistive load current of each segment is obtained as follows: in n segments of the distribution network, the th segment... i The resistance corresponding to each segment If j distribution transformers are subsequently connected, the formula for calculating the resistive load current flowing through the aforementioned section is:

[0101] in, For the first i The resistive load current of each segment, Let j be the rated capacity of the j-th transformer. Let j be the average load factor of the j-th transformer. For the first i The number of transformers in each section.

[0102] It should be noted that in the formula The rated line voltage of the distribution network must be used, and it must be consistent with the rated capacity of the transformer. The voltage level remains consistent; This is the number of transformers specific to the i-th segment and should not be confused with the number of transformers in other segments.

[0103] For example, suppose the first segment of a low-voltage distribution network connects two transformers: the first transformer... =100kVA It is 0.6, the second unit. =160kVA It is 0.55, rated line voltage If the voltage is 0.4kV, then the resistive load current of this section is:

[0104] S202. Add up the resistive load current of all segments to obtain the total resistive load current.

[0105] In some implementations, the formula for calculating the sum of resistive load currents is: ;in, This is the sum of the currents from the resistive load. For the first i The resistive load current of each segment.

[0106] It should be noted that the summation process must cover all n segments of the distribution network and cannot omit the current of any segment; the sum of currents must be consistent with the operating period of the subsequent loss calculation to ensure the time matching of the data; if a segment is an unloaded segment without a transformer, its resistive load current is taken as 0 and included in the summation.

[0107] In one possible implementation, combining Figure 1 ,like Figure 3 As shown, after S102, the method for calculating the network loss of a radial low-voltage distribution network based on the equivalent resistance method provided in this application embodiment further includes the following S301 and S302: S301. Calculate the power distribution loss of each segment using the line parameters and transformer data.

[0108] Among them, the resistance of each segment The resistance value needs to be calculated using the specifications of the conductor, which include: material, cross-section, and length.

[0109] In some implementations, the power distribution losses of each segment of the line... The formula for calculation is:

[0110] It should be noted that in the formula It needs to strictly correspond to the i-th segment; the "3" in the formula is the loss correction coefficient for a three-phase low-voltage distribution network. If it is a single-phase line, it needs to be adjusted to coefficient 1; when the unit of the calculation result is W, the unit needs to be kept consistent in subsequent statistics.

[0111] For example, if the resistive load current of the first segment The line resistance is 213.6A. If the resistance is 0.15Ω, then the power distribution loss of this section is: .

[0112] S302. Add up the line distribution losses of all segments to obtain the total line distribution loss.

[0113] In some implementations, the total line distribution loss The formula for calculation is:

[0114] It should be noted that the summation process must cover all n segments of the distribution network, and the losses of any segment cannot be omitted; the line distribution losses of each segment... They must correspond to the same runtime segment to ensure the time matching of the sum; if a segment is unloaded, its =0 must be included in the summation to ensure statistical integrity.

[0115] Based on the above technical solution, refined calculation of line loss per segment was achieved, avoiding the problem of ambiguous location of high-energy-consuming segments caused by the traditional general estimation of total line loss. This provides accurate single-segment energy consumption data for priority assessment of subsequent line energy-saving renovations. It also completes the statistical transition from single-segment loss to total line loss, providing an accurate loss data foundation for the subsequent derivation of the elementary equivalent resistance of the line, ensuring the accuracy of the equivalent resistance method in line loss calculation.

[0116] In implementation, each step of the method provided in this embodiment can be completed by integrated logic circuits in the processor or by instructions in software form. The steps of the method disclosed in the embodiments of this application can be directly manifested as being executed by a hardware processor, or being executed by a combination of hardware and software modules in the processor.

[0117] The processor in this application may include, but is not limited to, at least one of the following: a central processing unit (CPU), a microprocessor, a digital signal processor (DSP), a microcontroller unit (MCU), or an artificial intelligence processor, etc., which are various computing devices that run software. Each computing device may include one or more cores for executing software instructions to perform calculations or processing. The processor may be a separate semiconductor chip or integrated with other circuits into a single semiconductor chip. For example, it may be integrated with other circuits (such as encoding / decoding circuits, hardware acceleration circuits, or various bus and interface circuits) to form a SoC (System-on-a-Chip), or it may be integrated as a built-in processor within an ASIC. The ASIC with the integrated processor may be packaged separately or together with other circuits. In addition to the cores for executing software instructions to perform calculations or processing, the processor may further include necessary hardware accelerators, such as field-programmable gate arrays (FPGAs), PLDs (programmable logic devices), or logic circuits that implement dedicated logic operations.

[0118] The memory in the embodiments of this application may include at least one of the following types: read-only memory (ROM) or other types of static storage devices capable of storing static information and instructions; random access memory (RAM) or other types of dynamic storage devices capable of storing information and instructions; or electrically erasable programmable-only memory (EEPROM). In some scenarios, the memory may also be a compact disc read-only memory (CD-ROM) or other optical disc storage, optical disc storage (including compressed optical discs, laser discs, optical discs, digital universal optical discs, Blu-ray discs, etc.), magnetic disk storage media, or other magnetic storage devices, or any other medium capable of carrying or storing desired program code in the form of instructions or data structures and accessible by a computer, but is not limited thereto.

[0119] This application also provides a computer-readable storage medium including instructions that, when run on a computer, cause the computer to perform any of the methods described above.

[0120] This application also provides a computer program product containing instructions that, when run on a computer, cause the computer to perform any of the methods described above.

[0121] This application also provides a chip including a processor and an interface circuit. The interface circuit is coupled to the processor. The processor is used to run computer programs or instructions to implement the above-described method. The interface circuit is used to communicate with other modules outside the chip.

[0122] In the above embodiments, implementation can be achieved, in whole or in part, through software, hardware, firmware, or any combination thereof. When implemented using software programs, implementation can be, in whole or in part, in the form of a computer program product. This computer program product includes one or more computer instructions. When the computer program instructions are loaded and executed on a computer, all or part of the processes or functions described in the embodiments of this application are generated. The computer can be a general-purpose computer, a special-purpose computer, a computer network, or other programmable device. The computer instructions can be stored in a computer-readable storage medium or transmitted from one computer-readable storage medium to another. For example, computer instructions can be transmitted from one website, computer, server, or data center to another website, computer, server, or data center via wired (e.g., coaxial cable, fiber optic, digital subscriber line (DSL)) or wireless (e.g., infrared, wireless, microwave, etc.) means. The computer-readable storage medium can be any available medium accessible to a computer or a data storage device containing one or more servers, data centers, etc., that can be integrated with the medium. The available media can be magnetic media (e.g., floppy disks, hard disks, magnetic tapes), optical media (e.g., DVDs), or semiconductor media (e.g., solid-state disks (SSDs)).

[0123] Although this application has been described herein in conjunction with various embodiments, those skilled in the art, by reviewing the accompanying drawings, disclosure, and appended claims, will understand and implement other variations of the disclosed embodiments in carrying out the claimed application. In the claims, the word "comprising" does not exclude other components or steps, and "a" or "an" does not exclude multiple instances. A single processor or other unit can implement several functions listed in the claims. While different dependent claims may recite certain measures, this does not mean that these measures cannot be combined to produce good results.

[0124] Although this application has been described in conjunction with specific features and embodiments, it is obvious that various modifications and combinations can be made thereto without departing from the spirit and scope of this application. Accordingly, this specification and drawings are merely exemplary illustrations of this application as defined by the appended claims, and are considered to cover any and all modifications, variations, combinations, or equivalents within the scope of this application. Clearly, those skilled in the art can make various alterations and modifications to this application without departing from the spirit and scope of this application. Thus, if such modifications and modifications of this application fall within the scope of the claims of this application and their equivalents, this application is also intended to include such modifications and modifications.

Claims

1. A method for calculating losses in radial low-voltage distribution networks based on the equivalent resistance method, characterized in that, include: The power distribution network is divided into n segments, and the line parameters and transformer data of each segment are collected; wherein, the line parameters include resistive load current and line distribution loss; the transformer data includes: load current and load loss power; After calculating the elementary equivalent resistance of the line by summing the resistive load current of each segment and summing the line distribution loss, the equivalent resistance of the line is obtained by temperature correction. After calculating the elementary transformer equivalent resistance by summing the load current and summing the load loss power of each transformer segment, the transformer equivalent resistance is obtained by correcting it according to the three-phase unbalance. Calculate the total power loss of the distribution network during its operating time based on the equivalent resistance of the lines and transformers.

2. The method according to claim 1, characterized in that, The method for obtaining the equivalent resistance of the line includes: Based on temperature correction factor The equivalent resistance of the elementary circuit is corrected to obtain the equivalent resistance of the circuit; the formula for calculating the equivalent resistance of the circuit is: in, The equivalent resistance of the line. For elementary circuits, the equivalent resistance is... This represents the actual ambient temperature.

3. The method according to claim 1, characterized in that, The formula for calculating the equivalent resistance of the elementary transformer is: in, This is the equivalent resistance of an elementary transformer. This represents the total power loss due to the load. This represents the total load current of the transformer.

4. The method according to claim 3, characterized in that, The method for obtaining the equivalent resistance of the transformer includes: The equivalent resistance of the transformer is calculated using the elementary equivalent resistance and the total three-phase unbalance; the formula for calculating the equivalent resistance of the transformer is: in, The equivalent resistance of the transformer. This is the equivalent resistance of an elementary transformer. The total three-phase imbalance; The method for obtaining the total three-phase imbalance is as follows: obtain the transformer current of the phase with the largest phase imbalance. and the average value of the three-phase current of the transformer The total three-phase imbalance is calculated; the total three-phase imbalance The formula for calculation is: .

5. The method according to claim 1, characterized in that, The formula for calculating the total power loss of the distribution network is: in, The total power loss of the distribution network during the operating time period t. Let J be the no-load loss of the j-th distribution transformer in the distribution network. This is the equivalent current for distribution network losses.

6. The method according to claim 1, characterized in that, The formula for calculating the equivalent resistance of the elementary circuit is: in, For elementary circuits, the equivalent resistance is... This represents the total power distribution loss of the line. This represents the sum of the currents from the resistive load.

7. The method according to claim 1, characterized in that, The methods for obtaining the sum of the resistive load currents include: Calculate the resistive load current of each segment using the line parameters and transformer data of each segment; The sum of the resistive load currents of all segments is obtained by adding them together; the formula for calculating the sum of the resistive load currents is: in, This is the sum of the currents from the resistive load. For the first i The resistive load current of each segment; The method for obtaining the resistive load current of each segment is as follows: In n segments of the distribution network, the nth segment... i The resistance corresponding to each segment If j distribution transformers are subsequently connected, the formula for calculating the resistive load current flowing through the aforementioned section is: in, For the first i The resistive load current of each segment, Let j be the rated capacity of the j-th transformer. Let j be the average load factor of the j-th transformer. For the first i The number of transformers in each section.

8. The method according to claim 7, characterized in that, The formula for calculating the total power distribution loss of the line is: Calculate the power distribution loss of each segment by using the line parameters and transformer data of the segment; The total line distribution loss is obtained by adding up the distribution losses of all segments; the total line distribution loss... The formula for calculation is: The power distribution losses of each section of the line The formula for calculation is: .

9. The method according to claim 1, characterized in that, The method for obtaining the total load current of the transformer includes: Based on the transformer segment data, the load current of the j-th transformer is calculated, and the load currents of all transformers are summed to obtain the total load current of the transformers; the formula for calculating the total load current of the transformers is: in, This represents the total load current of the transformer. Let be the load current of the j-th transformer; The formula for calculating the load current of the j-th transformer is: in, Let J be the load current of the j-th transformer. Let j be the rated capacity of the j-th transformer. Let U be the average load factor of the j-th transformer. N This is the system's rated voltage.

10. The method according to claim 1, characterized in that, The formula for calculating the total load loss power is: in, This represents the total power loss due to the load. This refers to the steady-state short-circuit loss of the transformer. Let be the average load factor of the j-th transformer.

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