Method for calculating power factor of common connection point of AC metro power supply system

Abstract

The application discloses a kind of AC subway power supply system public connection point power factor calculation method, comprising the following steps: obtaining the line data required for calculating the power factor of the power supply system public connection point, electrical data;The mutual admittance Y de Between electrical node d and e is calculated, the self-admittance Y dd Of electrical node d is obtained, and the admittance matrix of AC subway power supply system node is obtained;The complex power of No.1 and No.2 electrical nodes is calculated, and the power factor of the public connection point of No.1 incoming line and No.2 incoming line is obtained.The application can provide basis for designers to configure the capacity of reactive power compensation device, provide decision support for system operation personnel to input compensation capacity and balance system reactive power, so as to better guarantee the safe and reliable operation of AC subway power supply system.

Description

Technical Field

[0001] This invention belongs to the field of power flow calculation for AC subway power supply systems, and specifically relates to a method for calculating the power factor at the point of common coupling of an AC subway power supply system. Background Technology

[0002] AC-powered subway systems offer advantages such as high train speeds, low track construction costs, and minimal stray current interference, making them a viable solution to the problems associated with DC-powered subways. The safe and reliable operation of the power supply system is crucial for ensuring the rapid and stable operation of locomotives, and reactive power balance has a significant impact on the operation of AC-powered subway power supply systems. Excessive reactive power can cause system voltage rise, leading to insulation damage and reduced service life of electrical equipment; increased system losses; and a decrease in the power factor at the point of common coupling. Therefore, it is necessary to address the reactive power imbalance problem in the power supply system by obtaining the power factor at the point of common coupling through power flow calculations, thereby understanding the reactive power level of the entire power supply system. This is of great significance for achieving the safe, stable, and reliable operation of the power supply system.

[0003] Currently, methods for obtaining the power factor at the point of common coupling (PCC) mainly include system data acquisition and power flow calculation. The advantage of acquiring PCC data through a Power System Monitoring and Data Acquisition (PSCADA) system is its ability to collect PCC data in real time, which helps in assessing the reactive power level of the power supply system. However, the location of the PCC varies between different city power departments. If the PCC is located at the outgoing line of a city substation, PSCADA will not be able to obtain the power factor for that PCC. Currently, most metro power flow calculation methods are applicable to DC metro systems, considering the impact of rectifier units and power feeder devices on the power supply system. However, AC metro systems do not contain such equipment and differ in structure and power supply system. Therefore, these methods are difficult to use for calculating the PCC power factor of AC metro systems. Summary of the Invention

[0004] To address the aforementioned problems, the present invention aims to provide a method for calculating the power factor at the point of common coupling (PCC) of an AC subway power supply system. This method can accurately calculate the PCC power factor of the AC subway power supply system, providing a basis for designers to configure the capacity of reactive power compensation devices, and providing decision support for system maintenance personnel to invest compensation capacity and balance system reactive power, thereby better ensuring the safe and reliable operation of the power supply system.

[0005] The technical solution adopted by this invention to achieve its objective is: a method for calculating the power factor at the point of common coupling of an AC subway power supply system, comprising the following steps:

[0006] A. Obtain data from the AC subway power supply system.

[0007] A1. Obtain the power supply system topology and assign a unique number to each electrical node in the power supply system as 1, 2, 3, ..., n according to the topology. Node 1 is the common connection point of incoming line 1, node 2 is the common connection point of incoming line 2, and n is the total number of electrical nodes in the power supply system.

[0008] A2. Obtain the basic parameters of transformers and cables in the power supply system.

[0009] The transformer parameters include the transformer's rated capacity S. N Short-circuit loss ΔP S Short-circuit voltage percentage U S %, No-load loss ΔP0, No-load current percentage I0%, Transformer primary side rated voltage U a The rated voltage U on the secondary side of the transformer b Transformer turns ratio C;

[0010] The cable parameters include the DC resistance per unit length r of cables with different nominal cross-sections. w Inductance per unit length L w Capacitance per unit length c w Where w = 1, 2, 3, ..., D, and D is the total number of different nominal cross-sectional areas of the cable; the highest voltage U of the cable during normal operation. max The minimum voltage U of the cable during normal operation min The length of each cable.

[0011] A3. Obtain power data of electrical nodes in the power supply system and the status of switches in the system.

[0012] The node power data includes the active power P of each electrical node. i and reactive power Q i Data, where i = 1, 2, 3, ..., n.

[0013] The status of the switches includes the closed / open status of the switches at both ends of each cable, and the Break status of the switches between nodes d and e (d = 1, 2, 3, ..., n, e = 1, 2, 3, ..., n, and d ≠ e). de The closing and opening status.

[0014] B. Establish the admittance matrix of the AC subway power supply system nodes.

[0015] B1. Write the node admittance matrix based on the power supply system topology.

[0016] B2. Calculate the mutual admittance Y between any two electrical nodes d and e. ded = 1, 2, 3, ..., n, e = 1, 2, 3, ..., n, and d ≠ e.

[0017] If there is no electrical connection between nodes d and e, the mutual admittance Y de Take 0.

[0018] If nodes d and e are connected by a cable, through R de =r w ·l de The equivalent resistance R of the cable was calculated. de Through X de =2πf·L w ·l de The equivalent inductive reactance X of the cable was calculated. de Through B de =2πf·c w ·l de The equivalent susceptance B of the cable was calculated. de , where l de The length of the cable connecting nodes d and e; f is the rated frequency, which is 50Hz; through The mutual admittance Y was calculated. de Where j is the imaginary unit; if the rated voltage at both ends of the cable is U a Then λ is taken as C; if the rated voltage at both ends of the cable is U b If λ is 1, then μ is 1; if both switches at both ends of the cable are closed, μ is 1; otherwise, μ is 10. 10 .

[0019] If nodes d and e are connected by a transformer, then... The equivalent resistance R of the transformer was calculated. T ,pass The equivalent inductive reactance X of the transformer was calculated. T ,pass The equivalent conductance G of the transformer was calculated. T ,pass The equivalent susceptance B of the transformer was calculated. T ;pass The mutual admittance Y was calculated. de .

[0020] If nodes d and e are connected by the switch Break de Connection, through The mutual admittance Y was calculated. de Among them, when the switch Break de When R is in the off state g X g Take 10 for each 10 When the switch breaks de When R is in a closed stateg X g All values ​​are 0.5.

[0021] B3, Through The self-admittance Y of electrical node d is calculated. dd .

[0022] If nodes d and e are connected only by the switch Break de Connection, Y d0 Take 0.

[0023] If nodes d and e are connected only by a transformer, or if nodes d and e are connected by both a transformer and a switch, then Break de Then Y d0 =G T -jB T .

[0024] If nodes d and e are connected only by a cable, or if nodes d and e are connected by a cable and a switch, then Break de ,but

[0025] If there is a cable and a transformer connecting nodes d and e, then

[0026] C. Calculate the power factor at the point of common coupling.

[0027] C1, take the initial voltage of any electrical node as...

[0028] C2. When i = 1 or 2, then take... Where k = 1, 2, 3, ..., k is the iteration number; when i ≠ 1 or 2, through The node voltage of the i-th node after the (k+1)-th iteration is obtained through iterative calculation, where m = 1, 2, 3, ..., n, and m ≠ i. yes The conjugate complex number; if or Pick

[0029] C3, when the node voltages at the (k+1)th and kth times... When, repeat step C2; when the node voltage of the (k+1)th and kth times... When the iteration calculation stops, the voltage of each electrical node is obtained.

[0030] C4. Substitute the calculated node voltages into... Obtain the complex power of each electrical node Where q = 1, 2, 3, ..., n, It is Yiq conjugate complex number, yes The conjugate of complex numbers.

[0031] C5. Take the complex power of nodes 1 and 2 respectively. and The real parts are P1 and P2, and the imaginary parts are Q1 and Q2; and according to The power factor at the point of common coupling of incoming line 1 was calculated, based on... The power factor of the common coupling point of incoming line No. 2 was calculated.

[0032] The beneficial technical effects of this invention are as follows:

[0033] 1. This invention takes into account the specific location of the common coupling point and can calculate the power factor of the common coupling point at a specified location more accurately.

[0034] 2. This invention takes into account the impact of power supply system network reconfiguration and can calculate the power factor of the point of common coupling under conditions such as cable out of operation and the closing and opening of switches connecting electrical nodes.

[0035] 3. This invention takes into account the structural differences between AC and DC metro systems, and the proposed power flow calculation method is applicable to calculating the power factor at the point of common coupling of AC metro power supply systems. Detailed Implementation

[0036] Example

[0037] A method for calculating the power factor at the point of common coupling in an AC subway power supply system, comprising the following steps:

[0038] A. Obtain data from the AC subway power supply system.

[0039] A1. Obtain the power supply system topology and assign a unique number to each electrical node in the power supply system as 1, 2, 3, ..., n according to the topology. Node 1 is the common connection point of incoming line 1, node 2 is the common connection point of incoming line 2, and n is the total number of electrical nodes in the power supply system.

[0040] A2. Obtain the basic parameters of transformers and cables in the power supply system.

[0041] The transformer parameters include the transformer's rated capacity S. N Short-circuit loss ΔP S Short-circuit voltage percentage U S %, No-load loss ΔP0, No-load current percentage I0%, Transformer primary side rated voltage U a The rated voltage U on the secondary side of the transformer b Transformer turns ratio C;

[0042] The cable parameters include the DC resistance per unit length r of cables with different nominal cross-sections. w Inductance per unit length L w Capacitance per unit length c w Where w = 1, 2, 3, ..., D, and D is the total number of different nominal cross-sectional areas of the cable; the highest voltage U of the cable during normal operation. max The minimum voltage U of the cable during normal operation min The length of each cable.

[0043] A3. Obtain power data of electrical nodes in the power supply system and the status of switches in the system.

[0044] The node power data includes the active power P of each electrical node. i and reactive power Q i Data, where i = 1, 2, 3, ..., n.

[0045] The status of the switches includes the closed / open status of the switches at both ends of each cable, and the Break status of the switches between nodes d and e (d = 1, 2, 3, ..., n, e = 1, 2, 3, ..., n, and d ≠ e). de The closing and opening status.

[0046] B. Establish the admittance matrix of the AC subway power supply system nodes.

[0047] B1. Write the node admittance matrix based on the power supply system topology.

[0048] B2. Calculate the mutual admittance Y between any two electrical nodes d and e. de d = 1, 2, 3, ..., n, e = 1, 2, 3, ..., n, and d ≠ e.

[0049] If there is no electrical connection between nodes d and e, the mutual admittance Y de Take 0.

[0050] If nodes d and e are connected by a cable, through R de =r w ·l de The equivalent resistance R of the cable was calculated. de Through X de =2πf·L w ·l de The equivalent inductive reactance X of the cable was calculated. de Through B de =2πf·c w ·l de The equivalent susceptance B of the cable was calculated. de , where lde The length of the cable connecting nodes d and e; f is the rated frequency, which is 50Hz; through The mutual admittance Y was calculated. de Where j is the imaginary unit; if the rated voltage at both ends of the cable is U a Then λ is taken as C; if the rated voltage at both ends of the cable is U b If λ is 1, then μ is 1; if both switches at both ends of the cable are closed, μ is 1; otherwise, μ is 10. 10 .

[0051] If nodes d and e are connected by a transformer, then... The equivalent resistance R of the transformer was calculated. T ,pass The equivalent inductive reactance X of the transformer was calculated. T ,pass The equivalent conductance G of the transformer was calculated. T ,pass The equivalent susceptance B of the transformer was calculated. T ;pass The mutual admittance Y was calculated. de .

[0052] If nodes d and e are connected by the switch Break de Connection, through The mutual admittance Y was calculated. de Among them, when the switch Break de When R is in the off state g X g Take 10 for each 10 When the switch breaks de When R is in a closed state g X g All values ​​are taken as 0.5.

[0053] B3, Through The self-admittance Y of electrical node d is calculated. dd .

[0054] If nodes d and e are connected only by the switch Break de Connection, Y d0 Take 0.

[0055] If nodes d and e are connected only by a transformer, or if nodes d and e are connected by both a transformer and a switch, then Break de Then Y d0 =G T -jB T .

[0056] If nodes d and e are connected only by a cable, or if nodes d and e are connected by a cable and a switch, then Breakde ,but

[0057] If there is a cable and a transformer connecting nodes d and e, then

[0058] C. Calculate the power factor at the point of common coupling.

[0059] C1, take the initial voltage of any electrical node as...

[0060] C2. When i = 1 or 2, then take... Where k = 1, 2, 3, ..., k is the iteration number; when i ≠ 1 or 2, through The node voltage of the i-th node after the (k+1)-th iteration is obtained through iterative calculation, where m = 1, 2, 3, ..., n, and m ≠ i. yes The conjugate complex number; if or Pick

[0061] C3, when the node voltages at the (k+1)th and kth times... When, repeat step C2; when the node voltage of the (k+1)th and kth times... When the iteration calculation stops, the voltage of each electrical node is obtained.

[0062] C4. Substitute the calculated node voltages into... Obtain the complex power of each electrical node Where q = 1, 2, 3, ..., n, It is Y iq conjugate complex number, yes The conjugate of complex numbers.

[0063] C5. Take the complex power of nodes 1 and 2 respectively. and The real parts are P1 and P2, and the imaginary parts are Q1 and Q2; and according to The power factor at the point of common coupling of incoming line 1 was calculated, based on... The power factor of the common coupling point of incoming line No. 2 was calculated.

Claims

1. A method for calculating the power factor at the point of common coupling (PCC) of an AC subway power supply system, characterized in that, Includes the following steps: A. Obtain data from the AC subway power supply system: A1. Obtain the power supply system topology and assign a unique number to each electrical node in the power supply system according to the topology, which is 1, 2, 3, ..., n. Node 1 is the common connection point of incoming line 1, node 2 is the common connection point of incoming line 2, and n is the total number of electrical nodes in the power supply system. A2. Obtain the basic parameters of transformers and cables in the power supply system; The transformer parameters include the transformer's rated capacity S. N Short-circuit loss ΔP S Short-circuit voltage percentage U S %, no-load loss ΔP0, no-load current percentage I0%, transformer primary side rated voltage U a The rated voltage U on the secondary side of the transformer b Transformer turns ratio C; The cable parameters include the DC resistance per unit length r of cables with different nominal cross-sections. w Inductance per unit length L w Capacitance per unit length c w Where w = 1, 2, 3, ..., D, and D is the total number of different nominal cross-sectional areas of the cable; the highest voltage U of the cable during normal operation. max The minimum voltage U of the cable during normal operation min The length of each cable; A3. Obtain power data of electrical nodes in the power supply system and the status of switches in the system; The node power data includes the active power P of each electrical node. i and reactive power Q i Data, where i = 1, 2, 3, ..., n; The status of the switches includes the closed and open status of the switches at both ends of each cable, and the Break status of the switches between nodes d and e (d=1, 2, 3, ..., n, e=1, 2, 3, ..., n, and d≠e). de The closing and opening status; B. Establish the admittance matrix of the AC subway power supply system nodes: B1. Write the node admittance matrix based on the power supply system topology. ; B2. Calculate the mutual admittance Y between any two electrical nodes d and e. de d = 1, 2, 3, ..., n, e = 1, 2, 3, ..., n, and d ≠ e; If there is no electrical connection between nodes d and e, the mutual admittance Y de Set to 0; If nodes d and e are connected by a cable, then... The equivalent resistance R of the cable was calculated. de ,pass The equivalent inductive reactance X of the cable was calculated. de ,pass The equivalent susceptance B of the cable was calculated. de , where l de The length of the cable connecting nodes d and e; f is the rated frequency, which is 50Hz; through The mutual admittance Y was calculated. de Where j is the imaginary unit; if the rated voltage at both ends of the cable is U a Then λ is taken as C; if the rated voltage at both ends of the cable is U b If λ is 1, then μ is 1; if both switches at both ends of the cable are closed, μ is 1; otherwise, μ is 1. ; If nodes d and e are connected by a transformer, then... The equivalent resistance R of the transformer was calculated. T ,pass The equivalent inductive reactance X of the transformer was calculated. T ,pass The equivalent conductance G of the transformer was calculated. T ,pass The equivalent susceptance B of the transformer was calculated. T ;pass The mutual admittance Y was calculated. de ; If nodes d and e are connected by the Break switch de Connection, through The mutual admittance Y was calculated. de Among them, when the switch Break de When R is in the off state g X g All When the switch breaks de When R is in a closed state g X g All values ​​are 0.5; B3, Through The self-admittance Y of electrical node d is calculated. dd ; If nodes d and e are connected only by the switch Break de Connection, Y d0 Set to 0; If nodes d and e are connected only by a transformer, or if nodes d and e are connected by both a transformer and a switch, then Break de ,but ; If nodes d and e are connected only by a cable, or if nodes d and e are connected by a cable and a switch, then Break de ,but ; If there is a cable and a transformer connecting nodes d and e, then ; C. Calculate the power factor at the point of common coupling: C1, take the initial voltage of any electrical node as... ; C2. When i = 1 or 2, then take... Where k = 1, 2, 3, ..., k is the iteration number; when i 1 or 2, through The node voltage of the i-th node after the (k+1)-th iteration is obtained by iterative calculation, where, Let be the self-admittance of electrical node i. Let m be the mutual admittance of electrical nodes i and m, where m = 1, 2, 3, ..., n, and m ≠ i. , Let these represent the node voltages of the m-th node after the k-th and (k+1)-th iterations, respectively. yes The conjugate complex number; if or ,Pick ; C3, when the node voltages at the (k+1)th and kth times... When, repeat step C2; when the node voltage of the (k+1)th and kth times... When the iteration calculation stops, the voltage of each electrical node is obtained; C4. Substitute the calculated node voltages into... The complex power of each electrical node is obtained. ,in, Let be the mutual admittances of electrical nodes i and q, where q = 1, 2, 3, ..., n. This represents the node voltage of the q-th node after the (k+1)-th iteration. It is Y iq conjugate complex number, yes The conjugate of complex numbers; C5. Take the complex power of nodes 1 and 2 respectively. and The real parts are P1 and P2, and the imaginary parts are Q1 and Q2; and according to The power factor at the point of common coupling of incoming line 1 was calculated, based on... The power factor of the common coupling point of incoming line No. 2 was calculated.

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