T-connection power transmission line fault traveling wave calculation method

Abstract

The invention relates to a T-connection power transmission line fault traveling wave calculation method. The method comprises the steps of node impedance matrix calculation, node current fault calculation, node voltage fault calculation and numerical Laplace inverse transformation. The method provided by the invention can accurately and conveniently calculate the fault traveling wave of the T-connection power transmission line, including the first wave head and a plurality of subsequent wave heads of the fault traveling wave, provides an efficient means for fault analysis of the T-connection power transmission line, and lays a theoretical foundation for relay protection research.

Description

Technical Field

[0001] The invention belongs to the technical field of fault location and relay protection in power systems, and particularly relates to a method for calculating fault traveling waves on a T-connected transmission line. Background Art

[0002] With the economic development of our country, the demand for electricity shows an increasing trend. In order to save the cost of equipment and improve the utilization efficiency of lines, high-voltage heavy-load three-terminal (T-type) or multi-terminal transmission lines with more than three terminals often appear. These lines are often connected to large power plants and large systems, and it is required to quickly cut off faults on the lines. The fault calculation of T-type transmission lines has also attracted more and more attention. Existing fault calculation methods for T-connected transmission lines usually only consider the calculation of fault steady-state quantities and do not involve the calculation of fault traveling waves. Existing fault traveling wave calculation methods usually adopt the grid method, but the fault traveling wave propagation process on a T-connected transmission line is extremely complex, and it is difficult for the grid method to calculate multiple waveheads of the fault traveling wave. To solve this problem, a precise calculation method for fault traveling waves on a T-connected transmission line has been invented. Summary of the Invention

[0003] The purpose of the invention is to provide a method for calculating fault traveling waves on a T-connected transmission line, which solves the problem that it is difficult for the grid method in the prior art to calculate multiple waveheads of the fault traveling wave.

[0004] The technical solution adopted by the invention to solve the above problems is to provide a method for calculating fault traveling waves on a T-connected transmission line, including the following steps:

[0005] Step 1: Input initial power supply parameters, T-connected transmission line parameters, and fault boundary conditions; where the initial power supply parameters are three-phase AC power supply parameters, including the positive-sequence equivalent resistance R of each phase voltage source eqn , the positive-sequence equivalent reactance X of each phase voltage source eqn , and the line voltage G of each phase voltage source n ; the T-connected transmission line parameters include the lengths l1, l2, and l3 of the three lines; the fault boundary conditions include the fault location, fault time, fault type, and transition resistance parameters, where the fault type includes three-phase ground short circuit, three-phase short circuit, two-phase ground short circuit, two-phase short circuit, and single-phase ground short circuit; the transition resistance parameters include y fa , y fb , y fc , y δ , y Σ , y fa , y fb , y fc , and y fa , y fb , y fc , y δ , and yΣ , y fa , y fb , y fc are the admittance values from the a-phase, b-phase, and c-phase of the fault point to the internal electrical node δ of the fault branch, respectively, and y Σ = y fa + y fb + y fc + y δ , y δ is the admittance value between the internal electrical node and the ground;

[0006] Step 2: Calculate the nodal admittance submatrix Y of the three-phase AC power supply connected to node n Gn , and calculate the nodal admittance matrix Y of the transmission line with its two ends connected to node j and node k respectively jk , where n, j, and k are node numbers; among them, the nodal admittance submatrix Y of the power supply connected to node n Gn is as follows:

[0007]

[0008] In the formula: R eqn is the positive-sequence equivalent resistance of power supply n, and X eqn is the positive-sequence equivalent reactance of power supply n, where the row and column numbers are both

[0009] The nodal admittance matrix Y with its two ends connected to node j and node k respectively jk is as follows:

[0010]

[0011] In the formula, Y jk(jx,jx) is the self-admittance of the x-phase of the transmission line at node j, Y jk(kx,kx) is the self-admittance of the x-phase of the transmission line at node k, Y jk(jx,jy) is the mutual admittance between the x-phase and y-phase of the transmission line at node j, Y jk(kx,ky) is the mutual admittance between the x-phase and y-phase of the transmission line at node k, (x, y = a, b, c, and x is not equal to y); Y jk(1,1) , Y jk(1,2) , Y jk(2,1) , Y jk(2,2) is the 3×3 submatrix of Y jk , where Y jk(1,1) represents the self-admittance of the transmission line at node j, and the row number and column number are both j, Y jk(1,2) represents the mutual admittance between node j and node k of the transmission line, and the row number and column number are j and k respectively; Y jk(2,1)Denotes the mutual admittance between nodes k and j of the transmission line, with the row number and column number being k and j respectively; Y jk(2,2) Denotes the self - admittance of the transmission line at node k, with both the row number and column number being k;

[0012] Step 3: Obtain the nodal admittance matrix Y of the entire system based on the nodal admittance matrix of each node's power source and the nodal impedance matrix of the transmission line fault And the nodal impedance matrix Z of the entire system fault ;

[0013] Add the nodal admittance sub - matrix of the power source with row number j and column number k and the nodal admittance sub - matrix of the transmission line to obtain the nodal admittance matrix Y fault The sub - matrix Y at the j - th row and k - th column fault (j,k), arrange all sub - matrices of Y fault in order to obtain the complete Y fault ;

[0014] Perform an inverse operation on Y fault to obtain the nodal impedance matrix Z fault ;

[0015] Step 4: Perform Norton equivalent on the non - fault area at the fault node to obtain the Norton equivalent admittance value Y eq of the non - fault branch and the Norton equivalent current source i eq ;

[0016] Among them, the Norton equivalent admittance value Y eq of the non - fault branch is:

[0017]

[0018] In the formula: Z fault(f，f) is the sub - matrix of Z fault with row number and column number , node f is the fault node; the Norton equivalent current source i eq is:

[0019] i eq = Y eq u eq

[0020] In the formula: u eq is the voltage value when the fault node f is operating normally;

[0021] Step 5: Calculate the nodal admittance matrix Y f ;

[0022]

[0023] Step 6: Further calculate the current Δ flowing through the faulty branch iff ;

[0024]

[0025] where: Δi ff is a 3×1 matrix;

[0026] Step 7: Calculate the voltage value u of any node in the system after the fault j (j is the node number) is:

[0027] u j = u j0 + Z fault (j,f)Δi ff

[0028] In the formula: u j0 is the voltage value when node j operates normally, Δi ff is a 3×1 matrix, Z fault(j，f) is the submatrix of Z fault with the row number being and the column number being , j is the node number;

[0029] Step 8: Perform the inverse Laplace transform on the calculated voltage and current values in the frequency domain to obtain the corresponding time-domain values. The inverse Laplace transform can be carried out using the following formula:

[0030]

[0031] Fun q = Fun[c + i(2q + 1)Δω]

[0032] c n = (2Δω / π)exp(ncΔt + iπn / N s )

[0033] In the formula: fun() represents the time-domain values of voltage and current, Fun() represents the frequency-domain values of voltage and current, Δt and 2Δω are the sampling intervals in the time domain and frequency domain respectively, N s represents the number of sampling times, n = 0, 1,..., N s - 1, c represents the damping coefficient, and σ represents the time window function.

[0034] Advantages of the present invention: The calculation method proposed by the present invention can accurately and conveniently calculate the traveling waves of T-connected transmission line faults, including the first wavefront and subsequent multiple wavefronts of the traveling waves, providing an efficient means for the fault analysis of T-connected transmission lines and laying a theoretical foundation for the research of relay protection. Description of the Drawings

[0035] Figure 1 It is the equivalent circuit diagram after the fault of the T-shaped transmission line of the present invention.

[0036] Figure 2 It is the pole structure diagram of the overhead line of the present invention.

[0037] Figure 3 It is the graph of the voltage change trend over time during single-phase ground short circuit in Embodiment 1 of the present invention.

[0038] Figure 4 It is the graph of the voltage change trend over time during three-phase ground short circuit in Embodiment 2 of the present invention. Detailed implementation manners

[0039] To make the objectives, technical solutions and beneficial effects of the present invention clearer, the following further describes the embodiments of the present invention in detail with reference to the drawings.

[0040] The present invention provides a traveling wave calculation method for T-connected transmission line faults, including the following steps:

[0041] Step 1: Input the initial power supply parameters, T-connected transmission line parameters and fault boundary conditions; wherein the initial power supply parameters are three-phase AC power supply parameters, including the positive-sequence equivalent resistance R of each phase voltage source eqn , the positive-sequence equivalent reactance X of each phase voltage source eqn , the line voltage G of each phase voltage source n ; the T-connected transmission line parameters include the lengths l1, l2, l3 of the three lines; the fault boundary conditions include the fault location, fault time, fault type, transition resistance parameters, wherein the fault type includes three-phase ground short circuit, three-phase short circuit, two-phase ground short circuit, two-phase short circuit, single-phase ground short circuit; the transition resistance parameters include y fa , y fb , y fc , y δ , y Σ , y fa , y fb , y fc , and y fa , y fb , y fc , y δ , y Σ , y fa , y fb , y fc are respectively the admittance values from the a-phase, b-phase, c-phase of the fault point to the internal electrical node δ of the fault branch, y Σ = y fa + y fb + y fc + y δ , yδ is the admittance value between the internal electrical node and the ground;

[0042] Step 2: Calculate the nodal admittance submatrix Y of the three-phase AC power supply connected to node n Gn , and calculate the nodal admittance matrix Y of the transmission line with its two ends connected to node j and node k respectively jk , where n, j, and k are node numbers; among them, the nodal admittance submatrix Y of the power supply connected to node n Gn is:

[0043]

[0044] In the formula: R eqn is the positive-sequence equivalent resistance of power supply n, X eqn is the positive-sequence equivalent reactance of power supply n, where the row and column numbers are both

[0045] The nodal admittance matrix Y with its two ends connected to node j and node k respectively jk is as follows:

[0046]

[0047] In the formula, Y jk(jx,jx) is the self-admittance of the x-phase of the transmission line at node j, Y jk(kx,kx) is the self-admittance of the x-phase of the transmission line at node k, Y jk(jx,jy) is the mutual admittance between the x-phase and y-phase of the transmission line at node j, Y jk(kx,ky) is the mutual admittance between the x-phase and y-phase of the transmission line at node k, (x, y = a, b, c, and x is not equal to y); Y jk(1,1) , Y jk(1,2) , Y jk(2,1) , Y jk(2,2) is the 3×3 dimensional submatrix of Y jk , where Y jk(1,1) represents the self-admittance of the transmission line at node j, the row number and column number are both j, Y jk(1,2) represents the mutual admittance between node j and node k of the transmission line, the row number and column number are j and k respectively; Y jk(2,1) represents the mutual admittance between node k and node j of the transmission line, the row number and column number are k and j respectively; Y jk(2,2) represents the self-admittance of the transmission line at node k, the row number and column number are both k;

[0048] Step 3: Obtain the nodal admittance matrix Y fault of the entire system and the nodal impedance matrix Z fault of the entire system;

[0049] Add the nodal admittance sub - matrix of the power source with row number j and column number k and the nodal admittance sub - matrix of the transmission line to obtain the nodal admittance matrix Y fault In the sub - matrix Y at the j - th row and k - th column fault (j,k), arrange all sub - matrices of Y fault in order to obtain the complete Y fault ;

[0050] Perform an inverse operation on Y fault to obtain the nodal impedance matrix Z fault ;

[0051] Step 4: Norton equivalent the non - fault area at the fault node to obtain the Norton equivalent admittance value Y eq of the non - fault branch and the Norton equivalent current source i eq ;

[0052] Among them, the Norton equivalent admittance value Y eq of the non - fault branch is:

[0053]

[0054] In the formula: Z fault(f，f) is the sub - matrix of Z fault with row number and column number , the node f is the fault node; the Norton equivalent current source i eq is:

[0055] i eq = Y eq u eq

[0056] In the formula: u eq is the voltage value of the fault node f during normal operation;

[0057] Step 5: Calculate the nodal admittance matrix Y f of the fault branch;

[0058]

[0059] Step 6: Further calculate the current △ iff flowing through the fault branch;

[0060]

[0061] Among them: Δi ff is a 3×1 matrix;

[0062] Step 7: Calculate the voltage value u j (j is the node number) of any node in the post - fault system as:

[0063] u j = u j0 + Z fault (j, f)Δi ff

[0064] Where: u j0 is the voltage value when node j operates normally, and Δi ff is a 3×1 matrix, and Z fault(j，f) is Z fault The row number is The column number is of the sub-matrix, and j is the node number;

[0065] Step Eight: Perform the inverse Laplace transform on the calculated voltage and current values in the frequency domain to obtain the corresponding time-domain values. The inverse Laplace transform can be carried out using the following formula:

[0066]

[0067] Fun q = Fun[c + i(2q + 1)Δω]

[0068] c n = (2Δω / π)exp(ncΔt + iπn / N s )

[0069] Where: fun() represents the time-domain values of voltage and current, Fun() represents the frequency-domain values of voltage and current, Δt and 2Δω are the sampling intervals in the time domain and frequency domain respectively, N s represents the number of sampling times, n = 0, 1,..., N s -1, c represents the damping coefficient, and σ represents the time window function.

[0070] Example 1:

[0071] This example takes a single-phase ground short circuit (phase A short circuit) as an example. The three-phase AC power supply parameters and the T-connected transmission line parameter data are specifically shown in Table 1 below:

[0072] Table 1 Three-phase AC power supply parameters and T-connected transmission line parameter data table

[0073]

[0074] Substitute the three-phase AC power supply parameters and the T-connected transmission line parameter data in Table 1 above into the PSCAD / EMTDC simulation software respectively. Using the calculation method of the present invention, the variation trend of the voltage of the T-connected transmission line after the fault with time is as Figure 3 shown. From Figure 3It can be seen that the trends of the two are the same and almost coincide, indicating that the proposed method has good calculation accuracy.

[0075] Embodiment 2:

[0076] Taking the three-phase ground short circuit (short circuit of phase A, phase B, and phase C) as an example, the parameters of the three-phase AC power supply and the parameters of the T-connected transmission line are specifically exemplified by the data listed in Table 2 below:

[0077] Table 2 Data Table of Three-Phase AC Power Supply Parameters and T-Connected Transmission Line Parameters

[0078]

[0079]

[0080] The parameters of the three-phase AC power supply and the parameters of the T-connected transmission line in Table 1 above are respectively substituted into the PSCAD / EMTDC simulation software, and the changing trend of the voltage of the T-connected transmission line over time after the fault is obtained by using the calculation method of the present invention as Figure 4 shown. It can be seen from Figure 4 that the trends of the two are the same and almost coincide, indicating that the proposed method has good calculation accuracy.

[0081] From the above, it can be known that the traveling wave calculation method for T-connected transmission line faults adopted by the present invention has good calculation accuracy, and can accurately and conveniently calculate the traveling waves of T-connected transmission line faults compared with the mesh network calculation, including the first wave head and many subsequent wave heads of the traveling waves, providing an efficient means for the fault analysis of T-connected transmission lines and laying a theoretical foundation for the research of relay protection.

[0082] The above shows and describes the basic principles, main features and advantages of the present invention. Those skilled in the art should understand that the present invention is not limited by the above embodiments. What is described in the above embodiments and the specification only illustrates the principles of the present invention. Without departing from the spirit and scope of the present invention, the present invention will have various changes and improvements, and these changes and improvements all fall within the scope of the claimed invention.

Claims

1. A method for calculating traveling waves of faults in T-connected transmission lines, characterized in that: The following steps are involved: Step 1: Input the initial power supply parameters, T-connected transmission line parameters and fault boundary conditions; the initial power supply parameters are the three-phase AC power supply parameters, including the positive sequence equivalent resistance R of each phase voltage source eqn , positive sequence equivalent reactance X of each phase voltage source eqn , the line voltage G of each phase voltage source n ; T-connected transmission line parameters include the lengths of the three lines l1, l2, l3; the fault boundary conditions include fault location, fault time, fault type, and transition resistance parameters, where the fault types include three-phase ground short circuit, three-phase short circuit, two-phase ground short circuit, two-phase short circuit, and single-phase ground short circuit; the transition resistance parameters include y fa ,y fb ,y fc ,y δ ,y Σ ,y fa ,y fb ,y fc , and y fa ,y fb ,y fc ,y δ ,y Σ ,y fa ,y fb ,y fc are the admittance values ​​from phase a, phase b, phase c of the fault point to the internal electrical node δ of the fault branch, y Σ =y fa +y fb +y fc +y δ ,y δ is the admittance value between the internal electrical node and the earth; Step 2: Calculate the node admittance submatrix Y of the three-phase AC power supply connected to node n Gn , calculate the node admittance matrix Y of the transmission line whose two ends are connected to node j and node k respectively jk , where n, j, k are node numbers; the node admittance submatrix Y of the power source connected to node n Gn for: Where: R eqn is the positive sequence equivalent resistance of power supply n, X eqn is the positive sequence equivalent reactance of power supply n, where the row and column numbers are The node admittance matrix Y is connected to node j and node k at both ends jk as follows: Where Y jk(jx,jx) is the self-admittance of the x-phase of the transmission line at node j, Y jk(kx,kx) is the self-admittance of the x-phase of the transmission line at node k, Y jk(jx,jy) is the mutual admittance between phase x and phase y of the transmission line at node j, Y jk(kx,ky) is the mutual admittance between phase x and phase y of the transmission line at node k, (x, y = a, b, c, and x is not equal to y); Y jk(1,1) ,Y jk(1,2) ,Y jk(2,1) ,Y jk(2,2) Yes jk A 3×3 dimensional submatrix, where Y jk(1,1) represents the self-admittance of the transmission line at node j, the row number and column number are both j, Y jk(1,2) represents the mutual admittance of the transmission line between nodes j and k, with row and column numbers j and k respectively; Y jk(2,1) represents the mutual admittance of the transmission line between nodes k and j, with row and column numbers k and j respectively; Y jk(2,2) represents the self-admittance of the transmission line at node k, and the row number and column number are both k; Step 3: According to the node admittance matrix of each node power source and the node impedance matrix of the transmission line, the node admittance matrix Y of the entire system is obtained. fault And the node resistance matrix Z of the entire system fault ; Add the node admittance submatrix of the power source with row number j and column number k to the node admittance submatrix of the transmission line to obtain the node admittance matrix Y fault The submatrix Y at row j and column k fault (j,k), Y fault All sub-matrices of are arranged in order to obtain the complete Y fault ; Y fault Perform the inverse operation to obtain the node impedance matrix Z fault ; Step 4: Perform Norton equivalent on the non-fault area at the fault node to obtain the Norton equivalent admittance value Y of the non-fault branch eq and Norton equivalent current source i eq ; The Norton equivalent admittance value Y of the non-fault branch is eq for: Where: Z fault(f，f) Z fault The row number is Column number is The sub-matrix of , node f is the fault node; Norton equivalent current source i eq for: i eq =Y eq u eq Where: u eq is the voltage value of the fault node f during normal operation; Step 5: Calculate the node admittance matrix Y of the fault branch f ; Step 6: Further calculate the current △ flowing through the fault branch iff ; Where: Δi ff is a 3×1 matrix; Step 7: Calculate the voltage value u of any node in the system after the fault j (j is the node number) is: u j =u j0 +Z fault (j,f)Δi ff Where: u j0 It is the voltage value of node j during normal operation, Δi ff is a 3×1 matrix, Z fault(j，f) Z fault The row number is Column number is The sub-matrix of , j is the node number; Step 8: Perform an inverse Laplace transform on the calculated voltage and current values ​​in the frequency domain to obtain the corresponding time domain values. The inverse Laplace transform can be performed using the following formula: Fun q =Fun[c+i(2q+1)Δω] c n =(2Δω / π)exp(ncΔt+iπn / N s ) Where: fun() represents the time domain value of voltage and current, Fun() represents the frequency domain value of voltage and current, Δt and 2Δω are the sampling intervals in the time domain and frequency domain respectively, N s Indicates the number of sampling times, n = 0, 1, ..., N s -1, c represents the damping coefficient, and σ represents the time window function.

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