CN105301492B - A kind of internal fuse protected qualification method in high-voltage capacitor unit - Google Patents
A kind of internal fuse protected qualification method in high-voltage capacitor unit Download PDFInfo
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Abstract
Internal fuse protected qualification method, concretely comprises the following steps in a kind of high-voltage capacitor unit：Establish equivalent circuit when some capacitor element punctures in capacitor unit；The equivalent circuit is solved, obtains the current expression for flowing through breakdown capacitor element branch；The energy for the Internal fuse connected according to current expression and the injection of Internal fuse resistance calculations with breakdown capacitor element；With reference to the parameter of whole capacitor device and the material of Internal fuse and size, judge that can the Internal fuse used in high-voltage capacitor unit reliably fuse.The present invention can the selected Internal fuse of accurate judgement high-voltage capacitor unit can reliably fuse, provide reliable guarantee for the suitable Internal fuse of high-voltage capacitor Unit selection.
Description
Technical Field
The invention relates to the technical field of power capacitors, in particular to a method for verifying the protection performance of an internal fuse in a high-voltage capacitor unit.
Background
In an electric power system, a capacitor device is composed of a plurality of capacitor units connected in series and parallel, and the inside of the capacitor unit is composed of a plurality of capacitor elements connected in series and parallel, each capacitor element being connected to a power supply through an internal fuse. The internal fuse is one of the main measures for protecting the interelectrode short circuit fault of the high-voltage capacitor, and has the function of fusing the internal fuse body by using short-circuit discharge energy when the capacitor element is subjected to the interelectrode short circuit, disconnecting the branch where the short-circuit capacitor element is located, and isolating the short-circuit capacitor element from a power supply, so that other intact capacitor elements can continue to operate.
In order to ensure that the internal fuse can be reliably blown when the capacitor element interpolar breaks down, a suitable size of the internal fuse must be selected. The internal fuse usually uses metallic copper as a main raw material, and the opening performance of the internal fuse is mainly influenced by the cross-sectional area and the length of the fuse body. When designing the size of the internal fuse, firstly, the discharge current flowing through the breakdown capacitor element when the capacitor element breaks down should be calculated, and then the energy injected into the internal fuse connected in series with the capacitor element is calculated according to the parameters of the discharge loop, so as to select the proper size of the internal fuse.
The size of the internal fuse is not suitable to be selected too large, and is not suitable to be selected too small, and the selection too large can cause that the internal fuse cannot be fused or cannot be fused reliably when the element is broken down, and even intermittent electric arcs are generated, so that the fault is enlarged; selection of too small a fuse can result in the internal fuse blowing or damaging the melt during normal operation or routine testing, resulting in the capacitor not operating properly. At present, the size of the fuse in the capacitor unit is mainly selected by a capacitor manufacturer according to experience, accurate theoretical calculation is lacked as support, and the situation of protection misoperation or fault expansion caused by unreasonable internal fuse configuration also occurs for many times in engineering.
Disclosure of Invention
The invention aims to overcome the defects in the prior art and provides a method for verifying the protection performance of an internal fuse for a high-voltage capacitor unit, so as to conveniently judge whether the internal fuse selected by the high-voltage capacitor unit is proper or not, further ensure that the internal fuse can be reliably fused when the high-voltage capacitor is broken down, and ensure the normal operation of a power system.
In order to solve the technical problems, the technical scheme adopted by the invention is as follows.
A method for verifying the protection performance of an internal fuse in a high-voltage capacitor unit comprises the following steps:
step 1) establishing an equivalent circuit when a certain capacitor element in a capacitor unit breaks down;
step 2) solving the equivalent circuit to obtain an expression of the current flowing through the branch of the breakdown capacitor element;
step 3) calculating the energy of the internal fuse which is injected into the capacitor element and is connected in series with the breakdown capacitor element according to the current expression and the internal fuse resistance;
and 4) judging whether the internal fuse used in the high-voltage capacitor unit can be reliably fused or not according to the result of the step 3) by combining the parameters of the whole capacitor device and the material and the size of the internal fuse.
According to the method for verifying the protection performance of the internal fuse in the high-voltage capacitor unit, when the medium-value circuit is established in the step 1), the high-voltage capacitor unit comprises a breakdown capacitor element, a capacitor element group connected with the breakdown capacitor element in parallel, a capacitor element group connected with the breakdown capacitor element in series and other capacitor units connected with the capacitor unit where the breakdown capacitor element is located in parallel.
According to the method for verifying the protection performance of the internal fuse in the high-voltage capacitor unit, when the equivalent circuit is solved in the step 2), a Laplace transformation and Laplace inverse transformation mode is adopted for solving, and a current expression i (t) is obtained through calculation according to a KVL law;
let M, n and M be the number of parallel capacitor elements in the capacitor device, the number of series capacitor elements in the capacitor unit and the number of parallel capacitor units, respectively, and the intermediate variable be
When b is 2 4ac > 0, the loop is in an over-damped state, the current i (t) does not oscillate, and the current expression is
When b is 2 -4ac&When the damping value is less than 0, the loop is in an underdamping state, the current i (t) attenuates and oscillates, and the current expression is
Wherein a = L (p + 1), b = R (p + 1), C = p/C;
l is the inductance of the capacitor element; r is a resistance value of the capacitor element; c is the capacitance of the capacitor element, U C Is the voltage across the capacitor element immediately before breakdown occurs; t is time.
In the method for verifying the protection performance of the internal fuse in the high-voltage capacitor unit, the energy injected into the internal fuse in the step 3) is calculated according to the following formula:
in the formula, W C The energy injected into the internal fuse, T, is the duration of the injection current i (T).
In the method for verifying the protection performance of the internal fuse in the high-voltage capacitor unit, the step 4) specifically comprises the following steps:
firstly, calculating the energy required by fusing a single internal fuse wire;
secondly, whether the internal fuse used in the high-voltage capacitor unit can be reliably blown is judged according to the requirement of the internal fuse blowing energy.
Due to the adoption of the technical scheme, the invention has the following technical progress effects:
the method can accurately judge whether the internal fuse selected by the high-voltage capacitor unit can be reliably fused, and provides reliable guarantee for selecting a proper internal fuse for the high-voltage capacitor unit. According to the invention, an equivalent circuit is established according to the actual condition of the capacitor element during breakdown, the element parameters are calculated by referring to the actual condition, the current flowing through the fault element is calculated by adopting Laplace transformation and inverse transformation, the actual operation condition is met, and the accuracy is high; the invention does not limit the number of elements, the number of element groups, the number of capacitor units, equivalent inductance, resistance and capacitance of capacitors, thereby being applicable to different situations and having strong adaptability.
Drawings
FIG. 1 is a flow chart of the present invention;
FIG. 2 is an equivalent circuit diagram of a capacitor element in breakdown;
fig. 3 is an equivalent circuit diagram when the capacitor element breaks down;
FIG. 4 is an equivalent circuit diagram after a pull transform;
FIG. 5 is an ATP simulation model of capacitor element breakdown;
FIG. 6 is a graph comparing an analytic solution of discharge current with a numerical solution when the device breaks down;
FIG. 7 is a graph of the voltage and current waveforms of the capacitor cell when a re-breakdown of the internal fuse rupture occurs;
FIG. 8 is a graph of the voltage and current waveforms of a capacitor cell during a blowing process at a line frequency continuous flow.
Detailed Description
The invention will be described in further detail below with reference to the figures and specific examples.
The method is used for judging the protection performance of the internal fuse in the high-voltage capacitor unit and avoiding fault expansion caused by improper selection of the internal fuse.
Step 1) determining an equivalent circuit of a capacitor device when a certain capacitor element in a capacitor unit breaks down.
In the invention, the setting capacitor device is composed of M capacitor units connected in series and parallel, and the inside of each capacitor unit is composed of M parallel capacitor elements and n series capacitor elements, and each capacitor element is connected with an internal fuse in series. The basic requirements for internal fuse opening and isolation are: when the capacitor element breaks down, the discharge current flowing through the fuse within the broken capacitor element should be sufficient to cause the internal fuse to rapidly melt and open, effectively isolating the failed element from the group of intact elements. From the operating conditions, this basic requirement should be ensured to be achieved within a certain voltage range: lower limit voltage U 1 Usually the lowest possible operating voltage of the capacitor, the upper voltage U 2 This corresponds to the maximum transient overvoltage that is allowed to be tolerated in operation. In general, for parallel capacitor U 1 ＝0.9U n ，U 2 ＝2.0U n (ii) a For series capacitor U 1 ＝0.5U n ，U 2 ＝2.3U n ，U n Is the nominal voltage of the capacitor unit.
When a certain capacitor element in the capacitor unit breaks down, other capacitor elements connected with the broken-down capacitor element in parallel can discharge to the broken-down point, and transient discharge current with high frequency, high amplitude and fast attenuation is generated. Because the breakdown of the element usually occurs near the voltage peak value, the power frequency current just crosses zero at the moment, and because the circuit outside the unit has inductance, the power frequency current can not be injected into the breakdown point instantly.
If the transient discharge current is large enough, the internal fuse wire can be quickly heated, melted and vaporized to form high-frequency current arc, and the high-frequency current zero-crossing arc quenching can be realized in a very short time, at the moment, residual voltage with a certain amplitude is remained on the complete component group participating in the discharge, and if the arc channel is not re-broken, the breakdown of the breakdown fault component can be successfully realized. Under the condition, the power frequency fault current cannot influence the switching-on and switching-off process, and the whole switching-on and switching-off process cannot exceed 1ms. This is an internal fuse action process that meets the requirements for opening.
If the transient discharge current is not enough to melt and open the internal fuse rapidly, the power frequency current enters an arc channel of the internal fuse to form a power frequency arc, although the current is not large, the arcing time is long, the energy injected by a power supply is large, the very limited arc extinguishing capability of the internal fuse hardly ensures that the power frequency current can be extinguished and opened when the power frequency current passes zero, the arc can be continuously burnt for several seconds or even longer, and the fault development is enlarged. The internal fuse opening performance at this time is not satisfactory.
The number m of parallel capacitor elements has a great influence on the fault discharge current and the opening effect of the internal fuse. The rated capacity and voltage of the capacitor element can usually only be changed within a small interval due to the influence of material properties and processes, and the energy injected into the branch of the fault element is mainly determined by the number m of parallel elements. When the number m of parallel connections is too small, the discharge energy is small, the internal fuse can not be quickly cut off, so that power frequency fault current is involved, and the probability of occurrence of re-breakdown is high. When m is too large, the discharge energy is too large, so that the arc channel is easily seriously carbonized and damaged in insulation, and the probability of heavy breakdown is relatively high. The heavy breakdown will generate higher overvoltage, which will seriously threaten the intact component, and even the internal fuse group explosion phenomenon will happen in serious cases. Therefore, when the internal fuse is opened, the occurrence of the re-breakdown is not allowed, and the occurrence of the re-breakdown means that the opening fails.
According to the internal structure characteristics of the capacitor unit, the following analysis and assumption are firstly carried out when the internal fuse discharge current is calculated:
1) Because the connection line between the parallel elements is mainly an internal fuse, the inductance of the capacitor element is very small, and the resistance R of the internal fuse is considered in calculation f I.e. the total resistance R of the branches of the capacitor element, the inductance L of the internal fuse f Is the total inductance L of the capacitor element branch.
2) Inductance L of capacitor unit outgoing line D Inductance L of internal fuse f Much smaller, and may be disregarded; the resistance and inductance of the lead between the capacitor units are very small and can be ignored.
3)R D Is the resistance of the lead tab of the capacitor element, R f And R D Damping effect in the circuit, small cross-sectional area of the internal fuse, and R f In contrast, R D Much smaller, R is calculated D Neglect it.
When one capacitor element breaks down, the other capacitor elements connected in parallel with the broken-down capacitor element will discharge to the broken-down point, and the other capacitor units connected in parallel with the capacitor unit where the broken-down capacitor element is located outside the capacitor unit will also discharge to the broken-down capacitor element, so that the equivalent circuit of the discharge loop can be determined as the circuit diagram shown in fig. 2.
In FIG. 2, C 1 Is a breakdown capacitor element, assuming that its capacitance value is C and its resistance and inductance value are R respectively 1 And L 1 At the moment of breakdown, the voltage is U C ，；C 2 Is a capacitor element group connected in parallel with a breakdown capacitor element, having a capacitance value of (m-1) C and a resistance and an inductance value of R 2 And L 2 ；C 31 Is a capacitor element group connected in series with a breakdown capacitor element, and C is C if the number of series-connected elements in the capacitor unit is n 31 Has a capacitance value of mC/(n-1), and a resistance and an inductance value of R 31 And L 31 ，C 1 Breakdown transient C 31 The voltage on is (n-1) U C ；C 32 Is the other capacitor unit connected in parallel with the capacitor unit where the breakdown capacitor element is located, and assuming that the parallel number of the capacitor units is M, C is 32 Has a capacitance value of (M-1) mC/n, and a resistance and an inductance value of R 32 And L 32 ，C 1 Breakdown time C 32 Voltage at nU C 。
And 2) solving the equivalent circuit to obtain a current expression flowing through a branch where the breakdown capacitor element is located.
The values of the equivalent elements in fig. 2 can be calculated based on the above assumptions, specifically: r 1 ＝R f ＝R，L 1 ＝L f ＝L；R 2 ＝R/(m-1)，L 2 ＝L/(m-1)；R 31 ＝R(n-1)/m，L 31 ＝L(n--1)/m；R 32 ＝nR/[m(M-1)]，L 32 ＝nL/[m(M-1)]。
C 1 At breakdown, C 2 Discharging it and, at the same time, C 31 And C 32 After series connection, also pair C 1 And (4) discharging. Due to C 31 And C 32 Is in the opposite direction, so C 31 And C 32 Total voltage after series connection is nU C -(n-1)U C ＝U C ，C 31 And C 32 Total capacitance C after series connection 3 ＝(M-1)mC/[n+(M-1)(n--1)]，C 31 And C 32 Total resistance R after series connection 3 ＝R[n+(n-1)(M-1)]/[m(M-1)]，C 31 And C 32 Total inductance L after series connection 3 ＝L[n+(n-1)(M-1)]/[m(M-1)]。
The circuit shown in fig. 2 can be simplified to fig. 3 based on the above analysis calculation.
The circuit shown in fig. 3 does not satisfy the superposition principle, and is difficult to solve directly through a differential equation, so that laplace transformation is adopted for solving, and the transformed circuit is shown in fig. 4, and equations are written according to the KVL law, and are shown in equations (1) and (2).
By integrating the formula (1) and the formula (2), the following can be obtained by solving:
r is to be 1 、R 2 、R 3 、L 1 、L 2 、L 3 、C 2 、C 3 The expression of (c) is simplified after being substituted into the above expression:
definition ofThen:
assuming a = L (p + 1), b = R (p + 1), and C = p/C, the inverse laplace transform is performed by using a partial decomposition equation:
when b is 2 When 4ac is greater than 0, the loop is in an over-damping state, the current i (t) does not oscillate, and the following can be solved:
when b is 2 -4ac&When the current is less than 0, the loop is in an underdamped state, the current i (t) attenuates and oscillates, and the following can be solved:
wherein: l is the inductance of the element; r is the resistance value of the element; c is a capacitance value, U C Is the voltage across the front element at breakdown; t is time.
In general, since the loop resistance is small and the loop is in an underdamped state, the current flowing through the internal fuse is ringing.
Step 3) calculating the energy W of the internal fuse connected in series with the breakdown capacitor element according to the current expression and the internal fuse resistance C Specifically, it is calculated according to the formula (5).
The current simulation calculation and the example analysis of the internal fuse discharge are carried out by taking the parallel capacitor unit BAM6.56-556-1W as an example in the step.
The BAM6.56-556-1W type capacitor cell has 3 series sections of capacitor elements, each series section having 18 parallel capacitor elements, i.e. m =18, n =3, the rated voltage U of the cell n =6.56kV, rated capacitance value C n ＝41.1μF。
The capacitor element rated voltage was 2.19kV, rated capacitance value C =6.85 μ F, and the size was 332mm × 161.3mm × 16.8mm. The material of the internal fuse is copper, and the resistivity of the copper at different temperatures is shown in table 1. The melting temperature of the red copper is 1083 ℃, the time of the transient discharge fusing process of the internal fuse is very short, and for the convenience of calculation, the average value of the resistivity of the internal fuse in the temperature range of 0-1000 ℃ is 4.792 multiplied by 10 -8 Ω · m was estimated. The diameter of the internal fuse is 0.5mm, the length of the internal fuse is 120mm, and the resistance R of a single internal fuse can be calculated f ＝0.12m×4.792×10 -8 Ω·m/[π(0.25×10 -3 m) 2 ]=29.3m Ω, internal fuse inductance L f ＝(4π×10 -7 H/m)×0.12[ln(2×0.12m/(0.25×10 -3 m))-0.75]/(2π)＝0.147μH。
TABLE 1 resistivity of red copper at different temperatures
| Temperature (. Degree.C.) | 0 | 200 | 400 | 600 | 800 | 1000 |
| Resistivity ρ × 10 8 (Ω·m) | 1.694 | 2.93 | 4.03 | 5.3 | 6.7 | 8.1 |
According to the operation experience, the capacitor element usually breaks down near the voltage peak, so the calculation assumes
In practical engineering, considering the limitation of blasting energy, the capacitor bank adopts fancy wiring, and the number of directly parallel connected capacitor units is not more than 2, namely M =2. And calculating by substitution to obtain: p =0.04854, a =1.5414 × 10 -7 ，b＝0.03072，c＝7.0867×10 3 ，b 2 -4ac＝-0.003425<, 0, so the loop is in an underdamped state and the internal fuse current i (t) oscillates and attenuates.
According to the parameters, the following formula (4) is substituted:
i(t)＝1.0559×10 5 e -99660t sin(1.8985×10 5 t) (6)
in order to verify the correctness of the equation (4), a simulation model is built by using EMTP-ATP software as shown in FIG. 5, a discharge current i (t) when one element breaks down is obtained through simulation and is compared with the equation (6), and the result is shown in FIG. 6. As can be seen from fig. 6, the equation (4) derived by the analytical method is consistent with the numerical solution obtained by the simulation software, and the correctness of the equation (4) is verified.
As can be seen from fig. 6, the short-circuit discharge can be completed within 0.1ms, i.e., T =0.1ms. This is thus obtained according to equations (5) and (6):
substituting the resistance R =29.3m omega of the single internal fuse into calculation to obtain W C ＝642.42J。
And 4) judging whether the internal fuse used in the high-voltage capacitor unit can be reliably fused or not according to the result of the step 3) by combining the parameters of the whole capacitor device and the material and the size of the internal fuse.
For the parameters of the capacitor bank and the internal fuses in this embodiment, if only the number of parallel elements is changed, the energy injected into the fuse in the failed element is related to the number of parallel elements as shown in table 2.
TABLE 2 energy of injected internal fuse for different number m of parallel elements
| Number of parallel connections m | 2 | 3 | 4 | 5 | 6 | 7 |
| Injection of internal fuse energy (J) | 26.71 | 61.41 | 98.38 | 136.26 | 174.59 | 213.19 |
| Number of parallel connections m | 8 | 9 | 10 | 11 | 12 | 13 |
| Injection of internal fuse energy (J) | 251.94 | 290.81 | 329.75 | 368.74 | 407.78 | 446.85 |
| Number of parallel connections m | 14 | 15 | 16 | 17 | 18 | 19 |
| Injection of internal fuse energy (J) | 485.94 | 525.05 | 564.18 | 603.33 | 642.42 | 681.64 |
As can be seen from table 2, the energy injected into the fuse in the faulty element increases as the number m of parallel elements increases. Therefore, for capacitor units with different capacities, the energy of the injected internal fuse is calculated according to the above calculation process, and the internal fuses with different specifications are selected according to the injection energy. The internal fuse is selected to be too thick or too thin, which affects the protection performance.
A specific process of determining whether the internal fuse used in the high voltage capacitor unit can be reliably blown is as follows.
First, the energy required to blow a single internal fuse is calculated.
The density of copper is 8.9g/cm 3 The melting temperature was 1083 ℃, the vaporization temperature was 2562 ℃, and the specific heat of copper was 0.39X 10 3 J/(kg. DEG C.), the volume of the single internal fuse wire can be calculated to be 0.024cm 3 The mass was 0.21g. The blowing process of the internal fuse under the transient discharge current is very short and can be approximately considered as an adiabatic process. Therefore, the amount of heat required to heat all of one internal fuse from room temperature to the vaporization temperature was calculated to be 208.19J. The copper has a latent heat of vaporization of 300.3kJ/mol and a molar volume of 7.11X 10 -6 m 3 The amount of heat required to completely vaporize one internal fuse can be calculated to be 1.01 kJ/mol. Therefore, the energy required for heating an internal fuse from normal temperature to complete vaporization is 208.19J +1.01kJ =1.22kJ.
In actual operation, the internal fuse wire is not required to be blown with energy more than 1.22kJ, and only a part of the internal fuse wire is blown. Considering the dispersion of the internal fuse performance and the sufficient length of the internal fuse fracture to withstand the corresponding voltage requirement, the injection energy is required to be 1.5 times of the heat 208.19J required for heating the internal fuse from the normal temperature to the vaporization temperature, so as to ensure the reliable fusing and isolation of the internal fuse, i.e. 1.5 × 208.19J =312.29J.
Then, whether the internal fuse used in the high-voltage capacitor unit can be reliably blown is determined according to the requirement of the internal fuse blowing voltage.
As can be seen from equation (4), the energy injected into the internal fuse is proportional to the square of the voltage. Therefore, if required at 0.9U n The lower internal fuse can be reliably blown, and the energy requirement of the internal fuse to be injected under rated voltage is more than 312.29J (1/0.9) 2 =385.54J, and table look-up 2 shows that the number of parallel elements m is required to be more than or equal to 12. In the present embodiment, the number of capacitor parallel elements m =18, so that the internal fuse can be reliably blown when the capacitor element breaks down.
To verify the theoretical calculation results, an internal fuse isolation test at 0.9 times rated voltage was performed with the same type of capacitor, i.e., a BAM type 6.56-556-1W capacitor cell. The internal fuse has a diameter of 0.5mm and a length of 120mm, the number m of parallel capacitor elements 3 connected in series and in parallel 18, and the rated capacitance C =6.85 μ F.
During the test, the capacitor unit is applied with alternating current 0.9 times of rated voltage, and the capacitor element is punctured by utilizing metal conduction to simulate the breakdown short circuit of the capacitor element. Since the internal fuse short discharge occurs inside the capacitor unit, the discharge current thereof cannot be directly measured. Therefore, in the test, only the voltage and current waveforms of the capacitor cells were measured, and the test was repeated a plurality of times to record the waveform in which the puncture occurred in the vicinity of the voltage peak. Test results show that the capacitor unit can fuse the internal fuse in the transient discharge process under 0.9 times of rated voltage, the phenomenon of heavy breakdown of the internal fuse can not occur, and the internal fuse can reliably act. This is consistent with the calculations of the present invention.
In this step, if required, 0.5U n The lower internal fuse can be reliably blown, and the energy requirement of the internal fuse to be injected under rated voltage is more than 312.29J (1/0.5) 2 =1249.16J, table 2 shows that the number m of parallel elements is required to be significantly greater than 18, and that the remaining energy may not be sufficient to ensure reliable fusing although the fuse may rise to the vaporization temperature in some sections.
To verify the theoretical calculation results, internal fuse isolation tests at 0.5 times rated voltage were performed with capacitor cells of type BAM 6.56-556-1W. The internal fuse has a diameter of 0.5mm and a length of 120mm, and the capacitor elements 3 are connected in series and 18, and the rated capacitance C =6.85 μ F. When breakdown occurred at 0.5 times the rated voltage peak, the energy injected into the fuse in the breakdown capacitor element was 642.42J × (0.5) 2 160.61J, which is close to 208.19J, which is the amount of heat required to heat all of one internal fuse from room temperature to the vaporization temperature, is in the vicinity of the critical value of the internal fuse.
During the test, the capacitor unit is applied with alternating current 0.5 times of rated voltage, and the capacitor element is punctured by utilizing metal conduction to simulate the breakdown short circuit of the capacitor element. Since the internal fuse short discharge occurs inside the capacitor unit, the discharge current thereof cannot be directly measured. Therefore, in the test, only the voltage and current waveforms of the capacitor cells were measured, and the test was repeated a plurality of times to record the waveform in which the puncture occurred in the vicinity of the voltage peak. Test results show that the internal fuse can be usually fused in the transient discharge process when the capacitor unit is at 0.5 times of rated voltage, but the phenomenon of fracture re-breakdown occurs for many times. The waveform of the capacitor unit when the typical internal fuse break is re-broken is shown in fig. 7, and meanwhile, the situation that the internal fuse needs to be blown out after several seconds of power-frequency follow current often occurs, and the typical waveform of the capacitor unit in the process of blowing out under the power-frequency follow current is shown in fig. 8.
According to the test result, when the test capacitor unit breaks down at 0.5 times of rated voltage, the internal fuse has a small fracture, heavy breakdown is easy to occur, the internal fuse can be blown off only by depending on the energy of power frequency follow current, and at the moment, the internal fuse does not play a role in reliably isolating the broken-down capacitor element, namely, the internal fuse cannot reliably act. The experimental conclusion is consistent with the theoretical calculation result.
The above is the calculation result of the internal fuse with the diameter of 0.5mm and the length of 120mm, if the size of the internal fuse changes, the calculation result also changes correspondingly, and the engineering can judge according to the method of the invention.
In this embodiment, the energy injected into the fuse in series with the failed element when the first capacitor element breaks down is calculated, and when there is any breakdown of the other capacitor elements, the number of parallel capacitor elements is reduced from that of the first element, but the breakdown voltage is increased. The energy is proportional to the square of the voltage, so it can be roughly assumed that the energy injected into the fuse in the failed element when the subsequent capacitor element breaks down is substantially the same as when the first capacitor element breaks down.
Claims (3)
1. A method for verifying the protection performance of an internal fuse in a high-voltage capacitor unit is characterized by comprising the following steps:
step 1) establishing an equivalent circuit when a certain capacitor element in a capacitor unit breaks down;
step 2) solving the equivalent circuit to obtain an expression of the current flowing through the branch of the breakdown capacitor element;
step 3) calculating the energy of the internal fuse which is injected into the capacitor element and is connected in series with the breakdown capacitor element according to the current expression and the internal fuse resistance;
step 4) judging whether the internal fuse used in the high-voltage capacitor unit can be reliably fused or not according to the result of the step 3) by combining the parameters of the whole capacitor device and the material and the size of the internal fuse;
when the equivalent circuit is established in the step 1), the equivalent circuit comprises a breakdown capacitor element, a capacitor element group connected with the breakdown capacitor element in parallel, a capacitor element group connected with the breakdown capacitor element in series and other capacitor units connected with the capacitor unit where the breakdown capacitor element is located in parallel;
solving by adopting a Laplace transformation and Laplace inverse transformation mode when the equivalent circuit is solved in the step 2), and calculating according to a KVL law to obtain a current expression i (t);
let M, n and M be the number of parallel capacitor elements in the capacitor device, the number of series capacitor elements in the capacitor unit and the number of parallel capacitor units, respectively, the intermediate variables are:
when b is 2 When 4ac is more than 0, the loop is in an over-damping state, the current i (t) does not oscillate, and the current expression is as follows:
when b is 2 -4ac&When the current value is 0, the loop is in an underdamped state, the current i (t) attenuates and oscillates, and the current expression is as follows:
wherein a = L (p + 1), b = R (p + 1), C = p/C;
l is the inductance of the capacitor element; r is a resistance value of the capacitor element; c is the capacitance of the capacitor element, U C Is the voltage across the capacitor element immediately before breakdown occurs; t is time.
2. The method for verifying the protection performance of an internal fuse in a high-voltage capacitor unit as claimed in claim 1, wherein the energy for injecting the internal fuse in step 3) is calculated according to the following formula:
in the formula, W C The energy injected into the internal fuse, T, is the duration of the injection current i (T).
3. The method for verifying the protection performance of the internal fuse in the high-voltage capacitor unit according to claim 2, wherein the step 4) specifically comprises the following steps:
firstly, calculating the energy required by fusing a single internal fuse wire;
secondly, whether the internal fuse used in the high-voltage capacitor unit can be reliably blown is judged according to the requirement of the internal fuse blowing energy.
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| CN107622168B (en) * | 2017-09-26 | 2019-06-28 | 华中科技大学 | A kind of DC fuse relevant parameter determines method |
| CN111880009B (en) * | 2020-07-10 | 2023-05-30 | 广东电网有限责任公司广州供电局 | Medium-voltage fuse melt resistance accurate measurement circuit and measurement method thereof |
| CN112083357A (en) * | 2020-08-11 | 2020-12-15 | 中国电力科学研究院有限公司 | Method for measuring fusing energy of fuse in high-voltage parallel capacitor |
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