CN105301492A - Protection performance verification method for internal fuses of high voltage capacitor unit - Google Patents

Abstract

The invention provides a protection performance verification method for internal fuses of a high voltage capacitor unit. The verification method comprises the steps: establishing an equivalent circuit when one capacitor element is broken down in a capacitor unit; solving the equivalent circuit, and acquiring the current expression flowing through a capacitor element subcircuit; according to the current expression and the internal fuse resistance, calculating the energy of the tandem internal fuses, wherein the energy is injected into the capacitor element and breaks down the capacitor element; and determining whether the internal fuses used in the high voltage capacitor unit can be reliably fused by means of combination of the parameters of the whole capacitor device and the material and the size of the internal fuses. The protection performance verification method for internal fuses of a high voltage capacitor unit can accurately determine whether the internal fuses selected by the high voltage capacitor unit can be reliably fused, and can provide reliable guarantee for selecting suitable internal fuses for the high voltage capacitor unit.

Description

A kind of internal fuse protected qualification method in high-voltage capacitor unit

Technical field

The present invention relates to power capacitor technical field, particularly relate to a kind of internal fuse protected qualification method in high-voltage capacitor unit.

Background technology

In electric system, capacitor device is made up of multiple capacitor unit connection in series-parallel, and capacitor unit inside is then made up of multiple capacitor element connection in series-parallel, and each capacitor element is connected with power supply by an Internal fuse.Internal fuse is one of major measure of high-voltage capacitor intereelectrode short-circuit emergency protection; the effect of Internal fuse is when capacitor element generation intereelectrode short-circuit; utilize short circuit dischange energy fusing Internal fuse body; disconnect shorted condenser element place branch road; by shorted condenser element and isolated from power, the capacitor element making other intact can continue to run.

In order to ensure that Internal fuse reliably can fuse when capacitor element interpolar punctures, suitable Internal fuse size must be selected.Internal fuse adopts metallic copper as main raw material(s) usually, and its break performance is mainly subject to the impact of Internal fuse cross-sectional area and melt length.During design Internal fuse size, first should calculate capacitor element when puncturing, flow through the discharge current of breakdown potential container component, then calculate the energy injecting series connection with it Internal fuse according to discharge loop parameter, select suitable Internal fuse size according to this.

The size of Internal fuse should not be selected too much, also should not select too small, and when selecting meeting too much to cause component breakdown, Internal fuse can not fuse or can not reliably fuse, and even produces intermittent electric arc, and causing trouble expands; Select too little meeting causes Internal fuse, when normal operation or routine test, fusing or damage melt occur, cause capacitor cisco unity malfunction.At present, capacitor unit Internal fuse size is rule of thumb selected primarily of capacitor producer, lacks accurate theory calculate as support, also repeatedly occurs because Internal fuse unreasonable allocation causes the situation of false protection or failure propagation in engineering.

Summary of the invention

The object of the invention is to overcome defect in prior art; a kind of method of calibration of protective value of high-voltage capacitor unit Internal fuse is provided; so that judge that whether the Internal fuse of high-voltage capacitor Unit selection is suitable; further guarantee high-voltage capacitor Internal fuse when puncturing can reliably fuse, and ensures that electric system normally runs.

For solving the problems of the technologies described above, the technical solution used in the present invention is as follows.

A kind of internal fuse protected qualification method in high-voltage capacitor unit, verification is carried out according to the following steps:

Step 1) set up equivalent circuit when some capacitor elements puncture in capacitor unit;

Step 2) solve this equivalent circuit, obtain the current expression flowing through breakdown potential container component branch road;

Step 3) energy of the Internal fuse of connecting with breakdown potential container component is injected according to current expression and Internal fuse resistance calculations;

Step 4) according to step 3) result, in conjunction with 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.

Internal fuse protected qualification method in above-mentioned high-voltage capacitor unit; step 1) medium value circuit is when setting up, comprise breakdown potential container component, with the capacitor element group of breakdown potential container component parallel connection, with the capacitor element group that breakdown potential container component is connected and other capacitor units in parallel with breakdown potential container component place capacitor unit.

Internal fuse protected qualification method in above-mentioned high-voltage capacitor unit, step 2) in adopt Laplace conversion and Laplace inverse transformation mode to solve when solving equivalent circuit, and calculate acquisition current expression i (t) according to KVL law;

If m, n and M are respectively the serial number of capacitor element and the capacitor unit number of parallel connection in the number in parallel of capacitor element in capacitor device, capacitor unit, intermediate variable is

$p=\frac{\left(\frac{1}{m-1}\right)\left(\frac{(M-1)(n-1)+n}{m(M-1)}\right)}{(\frac{1}{m-1}+\frac{(M-1)(n-1)+n}{m(M-1)})};$

Work as b ²during-4ac > 0, loop is in overdamping state, current i (t) nonoscillatory, and current expression is

$i\left(t\right)=\frac{U_C}{\sqrt{b^2-4ac}}{e}^{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}+\frac{U_C}{-\sqrt{b^2-4ac}}{e}^{\left(\frac{-b-\sqrt{b^2-4ac}}{2a}t\right)}---\left(3\right)$

Work as b ²during-4ac<0, loop is in underdamping state, current i (t) damped oscillation, and current expression is

$i\left(t\right)=\frac{U_C}{a\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}}}{e}^{-\frac{b}{2a}t}\sin \left(\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}}t\right)---\left(4\right)$

In formula, a=L (p+1), b=R (p+1), c=p/C;

L is the inductance value of capacitor element; R is the resistance value of capacitor element; C is the capacitance of capacitor element, U _cfor puncturing the voltage occurred on front momentary capacitance device element; T is the time.

Internal fuse protected qualification method in above-mentioned high-voltage capacitor unit, step 3) in inject Internal fuse energy calculate according to following formula:

$W_C=\underset{0}{\overset{T}{\&Integral;}}{i}^2\left(t\right)Rdt---\left(5\right)$

In formula, W _cinject the energy of Internal fuse, T is the duration of Injection Current i (t).

Internal fuse protected qualification method in above-mentioned high-voltage capacitor unit, step 4) specifically comprise the following steps:

First, the energy required for the fusing of single Internal fuse is calculated;

Secondly, according to Internal fuse fusing energy requirement, judge that can the Internal fuse used in high-voltage capacitor unit reliably fuse.

Owing to have employed above technical scheme, the technical progress effect acquired by the present invention is as follows:

The present invention accurately can judge that can the Internal fuse selected by high-voltage capacitor unit reliably fuse, and the Internal fuse suitable for high-voltage capacitor Unit selection provides reliable guarantee.Actual conditions when the present invention punctures according to capacitor element establish equivalent circuit and calculate component parameter with reference to actual conditions, and employing Laplace conversion and inverse transformation calculate the electric current that fault element flows through, realistic ruuning situation, and accuracy is high; And the present invention does not limit number of elements, element group number and capacitor unit quantity and capacitor equivalent inductance, resistance and capacitance, therefore can be applicable to different situations, strong adaptability.

Accompanying drawing explanation

Fig. 1 is process flow diagram of the present invention;

Fig. 2 is the equivalent circuit diagram of capacitor element when puncturing;

Fig. 3 is the equivalent circuit diagram of capacitor element when puncturing;

Fig. 4 is the equivalent circuit diagram after pull-type conversion;

Fig. 5 is the ATP realistic model of capacitor element when puncturing;

Discharge current analytic solution and numerical solution comparison diagram when Fig. 6 is component breakdown;

The voltage and current oscillogram of capacitor unit when Fig. 7 is Internal fuse fracture generation repeated breakdown;

Fig. 8 is the capacitor unit voltage and current oscillogram of fusing process under power frequency continued flow.

Embodiment

Below in conjunction with the drawings and specific embodiments, the present invention is further elaborated.

The present invention is used for judging the protective value of Internal fuse in high-voltage capacitor unit, and avoid the expansion of causing trouble because Internal fuse selection is improper, concrete grammar comprises the following steps.

Step 1) determine the equivalent circuit of some capacitor elements capacitor device when puncturing in a capacitor unit.

In the present invention, setting capacitor device is made up of M capacitor unit connection in series-parallel, and capacitor unit is inner to be then made up of the capacitor element of m parallel connection and n capacitor element of connecting, and each capacitor element is connected an Internal fuse.Internal fuse cut-offs and the basic demand isolated is: when capacitor element punctures, the discharge current flowing through breakdown potential container component Internal fuse should be enough to make Internal fuse melt rapidly and cut-off, and fault element and intact element group is effectively isolated.Consider from service condition, this basic demand should be guaranteed to realize in certain voltage range: lower voltage limit U ₁be generally the minimum operation voltage that capacitor may occur, upper voltage limit U ₂then correspond to maximum transition superpotential allowed to bear in running.Usually, for shnt capacitor U ₁=0.9U _n, U ₂=2.0U _n; For series capacitor U ₁=0.5U _n, U ₂=2.3U _n, U _nfor the rated voltage of capacitor unit.

When in capacitor unit, a certain capacitor element punctures, other capacitor elements in parallel with breakdown potential container component can discharge to breakdown point, produce the transient state discharge current that frequency is high, amplitude is high, decay fast.Because component breakdown usually occurs in voltage peaks, now power current just in time zero passage, and there is inductance due to unit external circuit, therefore power current can not inject breakdown point at once.

If transient state discharge current is enough large, Internal fuse is made to generate heat rapidly, melt, vaporize, form high-frequency current electric arc, and within the extremely short time, realize the blow-out of high-frequency current zero passage, now, the intact element group of participation electric discharge will remain with the residual voltage of certain amplitude, and arc channel repeated breakdown if there is not, namely successfully achieves cut-offfing of breakdown fault element.In this case, power frequency fault current can not affect interrupting process, and whole interrupting process can not more than 1ms.This meets the Internal fuse course of action cut-offfing requirement.

Cut-off if transient state discharge current is not enough to make Internal fuse melt rapidly, power current is caused to enter the arc channel of Internal fuse, form power frequency arc, although electric current is little, but the energy that the arc time is long, power supply injects is large, the very limited arc extinguishing ability of Internal fuse is difficult to ensure that power current necessarily can cut-off when zero passage in blow-out, electric arc likely the sustained combustion several seconds even more of a specified duration, causing trouble development expands.Internal fuse break performance is now undesirable.

The cut-off effect of number m in parallel to fault discharge electric current and Internal fuse of capacitor element has a great impact.By the impact of material properties and technique, the rated capacity of capacitor element and voltage can only change usually in very little interval, and the energy injecting fault element branch road determines primarily of parallel element number m.When number m in parallel is too small, discharge energy is little, and Internal fuse can not cut-off rapidly and cause power frequency fault current to be got involved, and the probability that repeated breakdown occurs is high.When m is excessive, discharge energy is too large, easily makes the serious carbonization of arc channel, insulation harm, and the probability that repeated breakdown occurs is also relatively large.Repeated breakdown will produce higher superpotential, forms serious threat, time serious, the quick-fried phenomenon of Internal fuse group even can occur to intact element.Therefore, when Internal fuse cut-offs, repeated breakdown does not allow to occur, and occurs that namely repeated breakdown means failure of interruption.

According to capacitor unit inner structure feature, first carry out following analysis and hypothesis when carrying out Internal fuse discharge current and calculating:

1) due to the mainly Internal fuse of the line between parallel element, the inductance of capacitor element self is very little, thinks the resistance R of Internal fuse during calculating _fit is exactly the inductance L of the all-in resistance R of capacitor element branch road, Internal fuse _fbe exactly the total inductance L of capacitor element branch road.

2) inductance L of capacitor unit extension line _dthan the inductance L of Internal fuse _fmuch little, can not consider; The resistance gone between capacitor unit and inductance are very little, all can ignore.

3) R _dthe resistance of capacitor unit lead wire, R _fand R _dplay damping action in circuit, the cross-sectional area of Internal fuse is very little, with R _fcompare, R _dmuch smaller, R during calculating _dignore.

When a capacitor element punctures, other capacitor elements in parallel with breakdown potential container component can discharge to breakdown point, other capacitor units that capacitor unit is outside, be in parallel with the capacitor unit at this breakdown potential container component place also can discharge to breakdown potential container component, therefore, the equivalent circuit of discharge loop can be defined as circuit diagram as shown in Figure 2.

In Fig. 2, C ₁be breakdown potential container component, suppose that its capacitance is C, its resistance and inductance value are respectively R ₁and L ₁, the voltage puncturing moment is U _c; C ₂be the capacitor element group in parallel with breakdown potential container component, its capacitance is (m-1) C, and its resistance and inductance value are respectively R ₂and L ₂; C ₃₁be the capacitor element group of connecting with breakdown potential container component, suppose that capacitor unit inner member serial number is n, then C ₃₁capacitance be mC/ (n-1), its resistance and inductance value are respectively R ₃₁and L ₃₁, C ₁puncture moment C ₃₁on voltage be (n-1) U _c; C ₃₂be other capacitor units in parallel with breakdown potential container component place capacitor unit, suppose that the number in parallel of capacitor unit is M, then C ₃₂capacitance be (M-1) mC/n, its resistance and inductance value are respectively R ₃₂and L ₃₂, C ₁c when puncturing ₃₂on voltage be nU _c.

Step 2) solve this equivalent circuit, obtain the current expression flowing through breakdown potential container component place branch road.

The value of each equivalence element in Fig. 2 can be calculated according to above-mentioned hypothesis, be specially: 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 ₁when puncturing, C ₂to its electric discharge, meanwhile, C ₃₁with C ₃₂also to C after series connection ₁electric discharge.Due to C ₃₁with C ₃₂voltage direction contrary, so C ₃₁with C ₃₂total voltage after series connection is nU _c-(n-1) U _c=U _c, C ₃₁with C ₃₂total capacitance C after series connection ₃=(M-1) mC/ [n+ (M-1) (n--1)], C ₃₁with C ₃₂all-in resistance R after series connection ₃=R [n+ (n-1) (M-1)]/[m (M-1)], C ₃₁with C ₃₂total inductance L after series connection ₃=L [n+ (n-1) (M-1)]/[m (M-1)].

According to above-mentioned analytical calculation, can be Fig. 3 by circuit reduction shown in Fig. 2.

Circuit shown in Fig. 3 does not meet superposition principle, directly comparatively difficult by differential equation, therefore adopt Laplace transform to solve, the circuit after conversion as shown in Figure 4, equation is write, shown in (1), (2) according to KVL law row.

$\&lsqb;I_2\left(s\right)+I_3\left(s\right)\&rsqb;(R_1+{\mathrm{sL}}_1)+I_2\left(s\right)(R_2+{\mathrm{sL}}_2+\frac{1}{{\mathrm{sC}}_2})=\frac{U_C}{s}---\left(1\right)$

$I_2\left(s\right)(R_2+{\mathrm{sL}}_2+\frac{1}{{\mathrm{sC}}_2})=I_3\left(s\right)(R_3+{\mathrm{sL}}_3+\frac{1}{{\mathrm{sC}}_3})---\left(2\right)$

Comprehensively (1) formula and (2) formula, through solving and can obtaining:

$I\left(s\right)=I_2\left(s\right)+I_3\left(s\right)=\frac{U_C}{S}/\&lsqb;(R_1+{\mathrm{sL}}_1)+\frac{(R_2+{\mathrm{sL}}_2+\frac{1}{{\mathrm{sC}}_2})(R_3+{\mathrm{sL}}_3+\frac{1}{{\mathrm{sC}}_3})}{(R_2+{\mathrm{sL}}_2+\frac{1}{{\mathrm{sC}}_2})+(R_3+{\mathrm{sL}}_3+\frac{1}{{\mathrm{sC}}_3})}\&rsqb;$

By R ₁, R ₂, R ₃, L ₁, L ₂, L ₃, C ₂, C ₃expression formula bring above formula into after abbreviation can obtain:

$I\left(s\right)=\frac{U_C}{s\&lsqb;(R+sL+\frac{1}{sC})\frac{\left(\frac{1}{m-1}\right)\left(\frac{(M-1)(n-1)+n}{m(M-1)}\right)}{(\frac{1}{m-1}+\frac{(M-1)(n-1)+n}{m(M-1)})}+(R+sL)\&rsqb;}$

Definition $p=\frac{\left(\frac{1}{m-1}\right)\left(\frac{(M-1)(n-1)+n}{m(M-1)}\right)}{(\frac{1}{m-1}+\frac{(M-1)(n-1)+n}{m(M-1)})},$ Then:

$I\left(s\right)=\frac{U_C}{s^{2}L(p+1)+sR(p+1)+\frac{p}{C}}$

If a=L (p+1), b=R (p+1), c=p/C, utilizes partial fraction expansion method to carry out inverse Laplace transform and obtains:

Work as b ²during-4ac > 0, loop is in overdamping state, current i (t) nonoscillatory, can solve:

$i\left(t\right)=\frac{U_C}{\sqrt{b^2-4ac}}{e}^{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}t\right)}+\frac{U_C}{-\sqrt{b^2-4ac}}{e}^{\left(\frac{-b-\sqrt{b^2-4ac}}{2a}t\right)}---\left(3\right)$

Work as b ²during-4ac<0, loop is in underdamping state, current i (t) damped oscillation, can solve:

$i\left(t\right)=\frac{U_C}{a\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}}}{e}^{-\frac{b}{2a}t}sin\left(\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}}t\right)---\left(4\right)$

Wherein: L is the inductance value of element; R is the resistance value of element; C is capacitance, U _cvoltage before occurring during for puncturing on element; T is the time.

Generally, because loop resistance is less, be in underdamping state, the electric current therefore flowing through Internal fuse is damped oscillation.

Step 3) according to current expression and Internal fuse resistance, calculate the energy W injecting the Internal fuse of connecting with breakdown potential container component _c, particular according to formula (5) calculates.

$W_C=\underset{0}{\overset{T}{\&Integral;}}{i}^2\left(t\right)Rdt---\left(5\right)$

This step is for shnt capacitor unit B AM6.56-556-1W, and the current simulations of carrying out Internal fuse electric discharge calculates and instance analysis.

BAM6.56-556-1W type capacitor unit has 3 capacitor element series block, and each series block has 18 shnt capacitor elements, i.e. m=18, n=3, the rated voltage U of unit _n=6.56kV, rated capacity value C _n=41.1 μ F.

Capacitor element rated voltage is 2.19kV, and rated capacity value C=6.85 μ F, is of a size of 332mm × 161.3mm × 16.8mm.The material of Internal fuse is red copper, and red copper resistivity is at different temperatures as shown in table 1.The temperature of fusion of red copper is 1083 DEG C, and the Internal fuse transient state electric discharge fusing process time is very short, for the ease of calculating, gets the mean value 4.792 × 10 of Internal fuse resistivity in 0 ~ 1000 DEG C of temperature range ^-8Ω m estimates.The diameter of Internal fuse is 0.5mm, and length is 120mm, can calculate the resistance R of single Internal fuse _f=0.12m × 4.792 × 10 ^-8Ω m/ [π (0.25 × 10 ^-3m) ²]=29.3m Ω, Internal fuse inductance L _f=(4 π × 10 ^-7h/m) × 0.12 [ln (2 × 0.12m/ (0.25 × 10 ^-3m))-0.75]/(2 π)=0.147 μ H.

Table 1 red copper resistivity at different temperatures

Temperature (DEG C)	0	200	400	600	800	1000
							Electricalresistivityρ × 10 8(Ω·m)	1.694	2.93	4.03	5.3	6.7	8.1

According to operating experience, capacitor element is all puncture at voltage peaks usually, hypothesis when therefore calculating $U_C=\sqrt{2}\&times;2.19kV=3.09kV.$

In Practical Project, consider blasting energy quantitative limitation, Capacitor banks adopts fancy wiring, and capacitor unit directly in parallel is no more than 2, i.e. M=2.Substitution calculates: p=0.04854, a=1.5414 × 10 ^-7, b=0.03072, c=7.0867 × 10 ³, b ²-4ac=-0.003425<0, so loop is in underdamping state, Internal fuse current i (t) oscillatory extinction.

According to above-mentioned parameter, substitute into formula (4) and obtain:

i(t)＝1.0559×10 ⁵e ^-99660tsin(1.8985×10 ⁵t)(6)

In order to the correctness of verification expression (4), EMTP-ATP software is utilized to build realistic model as shown in Figure 5, by emulating discharge current i (t) when obtaining a component breakdown, and compare with formula (6), result as shown in Figure 6.As seen from Figure 6, the formula (4) utilizing analytical method to derive is consistent with the numerical solution utilizing simulation software to obtain, and demonstrates the correctness of formula (4).

As shown in Figure 6, short circuit dischange can complete in 0.1ms, i.e. T=0.1ms.Therefore obtain according to formula (5) and formula (6):

$\begin{array}{c}{W}_C=\underset{0}{\overset{0.0001}{\&Integral;}}{i}^2\left(t\right)Rdt\\ =1.1149\&times;{10}^{10}R\underset{0}{\overset{0.0001}{\&Integral;}}{e}^{-2\&times;99660t}{\sin }^2(1.8985\&times;{10}^{5}t)dt\\ =21925.7R\end{array}$

The resistance R=29.3m Ω of single Internal fuse is substituted into and calculates, obtain W _c=642.42J.

Step 4) according to step 3) result, in conjunction with 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.

For the parameter of Capacitor banks in the present embodiment and Internal fuse, during iff change parallel element quantity, the relation injecting the energy of fault element Internal fuse and parallel element quantity is as shown in table 2.

Injection Internal fuse energy during table 2 different parallel element number m

Number m in parallel	2	3	4	5	6	7
							Inject Internal fuse energy (J)	26.71	61.41	98.38	136.26	174.59	213.19
Number m in parallel	8	9	10	11	12	13
							Inject Internal fuse energy (J)	251.94	290.81	329.75	368.74	407.78	446.85
Number m in parallel	14	15	16	17	18	19
							Inject Internal fuse energy (J)	485.94	525.05	564.18	603.33	642.42	681.64

As shown in Table 2, along with the energy of the increase injection fault element Internal fuse of parallel element quantity m increases.Therefore the capacitor unit for different capabilities according to above-mentioned computation process, should calculate the energy injecting Internal fuse, and selects the Internal fuse of different size according to Implantation Energy.Internal fuse was selected thick or meticulously all will affect its protective value.

Judge that the detailed process that can the Internal fuse used in high-voltage capacitor unit reliably fuse is as follows.

First, the energy required for the fusing of single Internal fuse is calculated.

The density of copper is 8.9g/cm ³, temperature of fusion is 1083 DEG C, and vapourizing temperature is 2562 DEG C, and the specific heat of copper is 0.39 × 10 ³j/ (kg DEG C), the volume that can calculate single Internal fuse is 0.024cm ³, quality is 0.21g.The fusing process of Internal fuse under transient state discharge current is very short, can be similar to and think adiabatic process.Therefore, can calculate the heat that an Internal fuse is all heated to vapourizing temperature needs by normal temperature is 208.19J.The latent heat of vaporization of copper is 300.3kJ/mol, and the molar volume of copper is 7.11 × 10 ^-6m ³/ mol, can calculate the heat needed of an Internal fuse will being vaporized completely is 1.01kJ.Therefore, an Internal fuse being heated to by normal temperature the energy needed of vaporizing completely is 208.19J+1.01kJ=1.22kJ.

In actual moving process, not necessarily require that energy is greater than 1.22kJ Internal fuse and just can fuses, as long as Internal fuse has part to fuse.Consider that Internal fuse performance has dispersiveness, and Internal fuse fracture should have enough length can tolerate corresponding voltage request, therefore, require that Implantation Energy should be Internal fuse and can ensure that Internal fuse reliably fuses and isolates, i.e. 1.5 × 208.19J=312.29J by normal temperature 1.5 times of being all heated to heat 208.19J required for vapourizing temperature.

Secondly, according to the requirement of Internal fuse breakdown voltage, judge that can the Internal fuse used in high-voltage capacitor unit reliably fuse.

As can be seen from formula (4), inject square being directly proportional of energy and the voltage of Internal fuse.Therefore, if required at 0.9U _nlower Internal fuse can reliably fuse, then the energy requirement injecting Internal fuse under rated voltage is greater than 312.29J × (1/0.9) ²=385.54J, tables look-up 2 known, requires parallel element number m>=12.And in the present embodiment, capacitor parallel element number m=18, therefore when capacitor element punctures, Internal fuse can reliably fuse.

In order to proof theory result of calculation, with the capacitor of same model, namely BAM6.56-556-1W type capacitor unit has carried out the Internal fuse isolation experiment under 0.9 times of rated voltage.Its Internal fuse diameter is 0.5mm, and length is 120mm, and capacitor element 3 goes here and there 18 also, and namely parallel element number m is 18, element rated capacity value C=6.85 μ F.

During test, apply interchange 0.9 times of rated voltage to capacitor unit, utilize metallic conduction to puncture for capacitor element, the short circuit of analog capacitor component breakdown.Because Internal fuse short circuit dischange occurs in capacitor unit inside, its discharge current cannot directly be measured.Therefore, only measure the voltage and current waveform of capacitor unit during test, repeatedly repeat test, record puncture occurs in the waveform of voltage peaks.Test findings shows, capacitor unit all can fuse Internal fuse under 0.9 times of rated voltage in transient state discharge process, the phenomenon of Internal fuse fracture repeated breakdown can not occur, belong to the situation of Internal fuse energy action message.This and result of calculation of the present invention are coincide.

In this step, if required at 0.5U _nlower Internal fuse can reliably fuse, then the energy requirement injecting Internal fuse under rated voltage is greater than 312.29J × (1/0.5) ²=1249.16J, table look-up 2 known, require that parallel element number m is obviously greater than 18, although now part Internal fuse can be elevated to vapourizing temperature, remaining energy may not necessarily ensure reliable fusing.

In order to proof theory result of calculation, carry out the Internal fuse isolation experiment under 0.5 times of rated voltage with BAM6.56-556-1W type capacitor unit.Its Internal fuse diameter is 0.5mm, and length is 120mm, and capacitor element 3 goes here and there 18 also, element rated capacity value C=6.85 μ F.When puncturing under 0.5 times of rated voltage peak value, according to calculating, the energy injecting breakdown potential container component Internal fuse is 642.42J × (0.5) ²=160.61J, close to by an Internal fuse by normal temperature be all heated to vapourizing temperature need heat 208.19J, be in Internal fuse fusing critical value near.

During test, apply interchange 0.5 times of rated voltage to capacitor unit, utilize metallic conduction to puncture for capacitor element, the short circuit of analog capacitor component breakdown.Because Internal fuse short circuit dischange occurs in capacitor unit inside, its discharge current cannot directly be measured.Therefore, only measure the voltage and current waveform of capacitor unit during test, repeatedly repeat test, record puncture occurs in the waveform of voltage peaks.Test findings shows, when capacitor unit is under 0.5 times of rated voltage, Internal fuse also can fuse usually in transient state discharge process, but repeatedly occurs resolving salty punch-through.During typical Internal fuse fracture repeated breakdown, the waveform of capacitor unit as shown in Figure 7, also often occur the situation needing the power frequency continued flow Internal fuse through reaching the several seconds to fuse, under power frequency continued flow, the capacitor unit typical waveform of fusing process as shown in Figure 8 simultaneously.

As seen from the experiment, when this test product capacitor unit punctures under 0.5 times of rated voltage, Internal fuse fracture is very little, easy generation repeated breakdown, and often need to rely on the energy of power frequency continued flow could to fuse Internal fuse, now, Internal fuse does not play the effect of reliable isolation breakdown capacitor element, and namely Internal fuse can not action message.Conclusion (of pressure testing) is consistent with the calculated results.

Above-mentioned is the result of calculation of diameter 0.5mm, length 120mm Internal fuse, if Internal fuse size changes, then result of calculation also respective change can occur, and can carry out judging according to method of the present invention in engineering.

The present embodiment only calculates the energy injecting Internal fuse of connecting with fault element when first capacitor element punctures, when there being other capacitor elements to puncture again, reduce to some extent during the first component breakdown of the number ratio of shnt capacitor element, but voltage breakdown increases to some extent.Energy is directly proportional to voltage squared, therefore, can think roughly subsequent capacitance device component breakdown time inject fault element Internal fuse energy substantially identical when puncturing with first capacitor element.

Claims (5)

1. an internal fuse protected qualification method in high-voltage capacitor unit, is characterized in that, specifically comprise the following steps:

Step 1) set up equivalent circuit when some capacitor elements puncture in capacitor unit;

Step 2) solve this equivalent circuit, obtain the current expression flowing through breakdown potential container component branch road;

Step 3) energy of the Internal fuse of connecting with breakdown potential container component is injected according to current expression and Internal fuse resistance calculations;

Step 4) according to step 3) result, in conjunction with 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.

2. internal fuse protected qualification method in high-voltage capacitor unit according to claim 1; it is characterized in that; step 1) medium value circuit is when setting up, comprise breakdown potential container component, with the capacitor element group of breakdown potential container component parallel connection, with the capacitor element group that breakdown potential container component is connected and other capacitor units in parallel with breakdown potential container component place capacitor unit.

3. internal fuse protected qualification method in high-voltage capacitor unit according to claim 2, it is characterized in that, step 2) in adopt Laplace conversion and Laplace inverse transformation mode to solve when solving equivalent circuit, and calculate acquisition current expression i (t) according to KVL law;

If m, n and M are respectively the serial number of capacitor element and the capacitor unit number of parallel connection in the number in parallel of capacitor element in capacitor device, capacitor unit, intermediate variable is

$p=\frac{\left(\frac{1}{m-1}\right)\left(\frac{(M-1)(n-1)+n}{m(M-1)}\right)}{(\frac{1}{m-1}+\frac{(M-1)(n-1)+n}{m(M-1)})};$

Work as b ²during-4ac > 0, loop is in overdamping state, current i (t) nonoscillatory, and current expression is

$i\left(t\right)=\frac{U_C}{\sqrt{b^2-4ac}}{e}^{\left(\frac{-b+\sqrt{b^2+4ac}}{2a}t\right)}+\frac{U_C}{-\sqrt{b^2-4ac}}{e}^{\left(\frac{-b-\sqrt{b^2-4ac}}{2a}t\right)}---\left(3\right)$

Work as b ²during-4ac<0, loop is in underdamping state, current i (t) damped oscillation, and current expression is

$i\left(t\right)=\frac{U_C}{a\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}}}{e}^{-\frac{b}{2a}t}\sin \left(\sqrt{\frac{c}{a}-\frac{b^2}{4a^2}t}\right)---\left(4\right)$

In formula, a=L (p+1), b=R (p+1), c=p/C;

L is the inductance value of capacitor element; R is the resistance value of capacitor element; C is the capacitance of capacitor element, U _cfor puncturing the voltage occurred on front momentary capacitance device element; T is the time.

4. internal fuse protected qualification method in high-voltage capacitor unit according to claim 3, is characterized in that, step 3) in inject the energy of Internal fuse and calculate according to following formula:

$W_C=\underset{0}{\overset{T}{\&Integral;}}{i}^2\left(t\right)Rdt---\left(5\right)$

In formula, W _cinject the energy of Internal fuse, T is the duration of Injection Current i (t).

5. internal fuse protected qualification method in high-voltage capacitor unit according to claim 4, is characterized in that, step 4) specifically comprise the following steps:

First, the energy required for the fusing of single Internal fuse is calculated;

Secondly, according to Internal fuse fusing energy requirement, judge that can the Internal fuse used in high-voltage capacitor unit reliably fuse.