CN109245048A - It is a kind of to send out the differential protecting method become suitable for double-fed fan motor field - Google Patents
It is a kind of to send out the differential protecting method become suitable for double-fed fan motor field Download PDFInfo
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- H—ELECTRICITY
- H02—GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
- H02H—EMERGENCY PROTECTIVE CIRCUIT ARRANGEMENTS
- H02H7/00—Emergency protective circuit arrangements specially adapted for specific types of electric machines or apparatus or for sectionalised protection of cable or line systems, and effecting automatic switching in the event of an undesired change from normal working conditions
- H02H7/04—Emergency protective circuit arrangements specially adapted for specific types of electric machines or apparatus or for sectionalised protection of cable or line systems, and effecting automatic switching in the event of an undesired change from normal working conditions for transformers
- H02H7/045—Differential protection of transformers
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Abstract
The differential protecting method become is sent out suitable for double-fed fan motor field the present invention relates to a kind of, belongs to Relay Protection Technology in Power System field.The invention includes the following steps: whether occurring first with difference of phase currents Judging fault；When not having failure, protection does not start, and when a failure occurs it, protection starting carries out three-stage percentage differential, excitation surge current locking, differential quick-break element movement logic judgment.When any one mutually differential quick-break action value is greater than differential quick-break setting valve, protection act；When the variance that any one phase three-stage percentage differential meets the magnetizing inductance in operation condition and certain time interval is less than setting valve, protection act.Otherwise, protection is failure to actuate.The differential protecting method can solve double-fed fan motor field and send out when becoming internal fault because of second harmonic larger the problem of causing biased differential protection to be blocked, and can effectively improve the reliability that transformer differential protection is sent out in double-fed fan motor field, work well.
Description
Technical field
The differential protecting method become is sent out suitable for double-fed fan motor field the present invention relates to a kind of, belongs to electric system relay guarantor
Protect technical field.
Background technique
It increases substantially with the development of the global economy with energy-output ratio, between the reserves of the energy, production and use
Contradiction become increasingly conspicuous, become current countries in the world one of major issue anxious to be resolved.It therefore, is solution energy crisis, ring
The problems such as border is polluted, the research and development of the new energy such as wind energy, solar energy have become the very urgent demand of current mankind.Wherein, wind
It can be a kind of energy that cleaning is forever continuous, compared with traditional energy, wind-power electricity generation, which has, does not depend on extra power, without fuel price
Risk, cost of electricity-generating are stable, do not have the Environmental costs advantages such as carbon emission；Compared with solar energy, tide energy, the Industry Foundation of wind energy
Preferably, economic advantages are the most obvious, and not big environment influences；Moreover, available wind energy distribution is very wide in global range
It is general.Due to these unique advantages that wind-power electricity generation has, it is made to be increasingly becoming the important set of many National Sustainable Development Strategies
It is rapidly developed at part, and in countries in the world.
Wind power generating set type is more, and double-fed blower relies on its speed-regulating range width, small to excitation converter capacity requirement
The advantages that, become one of the mainstream model of MW class wind turbine group.As power grid is to the grid-connected performance requirement of Wind turbines
It is continuously improved, double-fed blower also exposes the drawbacks such as the ability of absorbing impact is weak in short-circuit of machine's port failure.Short circuit malfunction meeting
The high current and high voltage for causing Wind turbines rotor-side, cause Wind turbines off-grid, and it is unfavorable to bring to grid stability.Double-fed
Formula Wind turbines generally use Crowbar protection circuit to realize low-voltage crossing, and the phases-time is of short duration, and waveform is complicated, wind turbine
The short circuit current that the short circuit current and conventional synchronization motor that group provides provide has apparent difference, and fault characteristic is generator
Itself and convertor controls and the comprehensive of Preservation tactics respond.
THE WIND ENERGY RESOURCES IN CHINA Relatively centralized, wind power plant are mostly extensive centralization access, and electric energy is sent out by wind power plant to be become
Depressor (110kV/35kV, 330kV/110kV) is sent outside to system.Wind power plant is sent out and is become on transformer using conventional electric power at present
Depressor protection, principle be based on bilateral or three sides before and after failure voltage, electric current frequency having the same, utilize current differential
Principle differentiates internal fault external fault.Since the control technology and grid-connected mode of wind-driven generator are different from conventional synchronous generator, wind
Access area power grid is concentrated in power power generation, and there are significant differences in the electro-magnetic transient characteristic during failure, are based on conventional electric power system
The relay protection action performance of fault transient response characteristic not can guarantee.
Summary of the invention
The differential protection side become is sent out suitable for double-fed fan motor field the technical problem to be solved in the present invention is to provide a kind of
Method, the present invention have derived magnetizing inductance calculation formula according to transformer equivalent circuit diagram, have calculated the excitation in certain time interval
The variance of inductance formulates criterion according to the numerical characteristics of magnetizing inductance in varied situations, to judge that transformer is to encourage
Magnetic shoves or breaks down, and in conjunction with three-stage percentage differential and differential quick-break element, constitutes double-fed fan motor field and sends out the difference become
Dynamic protection scheme.
The technical solution adopted by the present invention is that: it is a kind of to send out the differential protecting method become suitable for double-fed fan motor field, first
Whether occurred using phase current mutation principle Judging fault；When not having failure, protection does not start, when a failure occurs it,
Protection starting carries out three-stage percentage differential, excitation surge current locking, differential quick-break element movement logic judgment, when any one phase
When differential quick-break action value is greater than differential quick-break setting valve, protection act；Any one phase three-stage percentage differential satisfaction acts item
When the variance of magnetizing inductance in part and certain time interval is less than setting valve, protection act.Otherwise, protection is failure to actuate.
Specific step is as follows for the differential protecting method that the double-fed fan motor field submitting becomes:
Step1, judge whether to break down according to difference of phase currents, as shown in formula (1)；When any in A, B, C three-phase
One phase difference of phase currentsGreater than differential protection initiation value IcdWhen, it is judged as and breaks down,
Protection starting；Conversely, being judged as fault-free.
In formula,The difference of phase currents of respectively k-th sampled point A, B, C phase,The current value of respectively k-th sampled point A, B, C phase, Point
Not Wei previous cycle A, B, C phase current value,Respectively the first two period A,
B, the current value of C phase, N are the sampling number of a power frequency period, and k is k-th of sampled point, IcdFor differential protection initiation value, kk
For safety factor, kbFor load tap changer coefficient, kTAFor the error coefficient of current transformer TA, INFor high voltage side of transformer
Rated current；
Step2, judge failure occur after, differential protection starting, differential protection logic include three parts: percentage differential member
Part, shove blocking element and differential quick-break element；
(1) three-stage percentage differential
In formula, IopFor differential protection action current, Iop.minFor minimum working current, Ires.minFor the minimum restraint current,
IresFor stalling current, InFor Circuit Fault on Secondary Transformer rated current, k1For percentage differential coefficient；
A, the difference current of B, C phase and stalling current such as formula (3) and formula (4) are shown；
In formula, IAop、IBop、ICopRespectively transformer A, B, C phase difference current, IAres、IBres、ICresRespectively transformer
A, B, C phase stalling current, iA、iB、iCRespectively transformer primary side A, B, C phase current, ia、ib、icRespectively transformer secondary
Side A, B, C phase line current；
The difference current of A, B, C phase and stalling current in formula (3) and formula (4) are brought into respectively in formula (2), when in three-phase
When a phase difference current of anticipating meets operation condition, three-stage percentage differential element movement；
(2) it shoves locking
1) according to the three-phase transformer T shape equivalent circuit diagram (see attached drawing 1) of Y/ Δ wiring, three-phase loop equation is formed, such as
Shown in formula (5)；
In formula (5), uA、uB、uCRespectively transformer Y shape side A, B, C phase voltage,Respectively transformer Δ side
A, B, C phase winding electric current, LA、LB、LCThe respectively side transformer Y A, B, C phase leakage inductance, RA、RB、RCRespectively transformer A, B, C phase
Resistance, RmA、RmB、RmCRespectively transformer A, B, C phase excitation resistance, LmA、LmB、LmCRespectively transformer A, B, C phase excitation electricity
Sense, ipFor the circulation in Δ winding, t is the time；
2)LA、RA、RmAIt is much smaller than LmA, LB、RB、RmBIt is much smaller than LmB, LC、RC、RmCIt is much smaller than LmC, therefore formula (5) can
Approximate representation is formula (6)；
3) i is eliminated using formula (6)p, as shown in formula (7)；
4) transformer Δ side A, B, C phase winding electric current in formula (7) is soughtAs shown in formula (8)-(9)；
Solution formula (8) obtains Δ side A, B, C phase winding electric currentAs shown in formula (9)；
5) formula (9) is brought into formula (7), as shown in formula (10)；
6) it enablesIt is right
Formula (10) carries out discretization, as shown in formula (11)；
In formula, uA(k)、uB(k)、uCIt (k) is respectively k-th of the side sampled point transformer Y A, B, C phase voltage, uA(k+1)、uB
(k+1)、uCIt (k+1) is respectively+1 side sampled point transformer Y A, B, C phase voltage of kth, Δ T is+1 sampled point of kth and kth
The time difference of a sampled point, i'abThe sum of the side Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcFor the side Y B, C two-phase
The sum of current difference and Δ side b, c biphase current difference, i 'caFor the sum of the side Y C, A biphase current difference and Δ side c, a biphase current difference,
i′abIt (k) is the sum of k-th of the side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k) is k-th of sampling
The sum of the side point Y B, C biphase current difference and Δ side b, c biphase current difference, i'ca(k) poor for k-th of the side sampled point Y C, A biphase current
The sum of with Δ side c, a biphase current difference, i'abIt (k+1) is+1 side sampled point Y A, B biphase current difference of kth and Δ side a, b two-phase
The sum of current difference, i 'bcIt (k+1) is the sum of+1 side sampled point Y B, C biphase current difference of kth and Δ side b, c biphase current difference,
i'caIt (k+1) is the sum of+1 side sampled point Y C, A biphase current difference of kth and Δ side c, a biphase current difference, i'abIt (k-1) is the
The sum of the k-1 side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k-1) is -1 side sampled point Y of kth
B, the sum of C biphase current difference and Δ side b, c biphase current difference, i'ca(k-1) poor for -1 side sampled point Y C, A biphase current of kth
The sum of with Δ side c, a biphase current difference, i'abIt (k+2) is+2 side sampled point Y A, B biphase current differences of kth and Δ side a, b two-phase
The sum of current difference, i'bcIt (k+2) is the sum of+2 side sampled point Y B, C biphase current differences of kth and Δ side b, c biphase current difference,
i′caIt (k+2) is respectively the sum of+2 side sampled point Y C, A biphase current differences of kth and Δ side c, a biphase current difference；
7) magnetizing inductance that formula (11) respectively obtain A, B, C phase is solved, as shown in formula (12)；
In formula, Δi'abIt (k) is k-th of sampled point i'abDiscrete value, Δ i'abIt (k+1) is kth+1
Sampled point i'abDiscrete value, Δ i 'bcIt (k) is k-th of sampled point i 'bcDiscrete value, Δ i 'bcIt (k+1) is+1 sampled point i ' of kthbc
Discrete value, Δ i'caIt (k) is k-th of sampled point i 'caDiscrete value, i'caIt (k+1) is+1 sampled point i ' of kthcaDiscrete value.
8) variance for solving the magnetizing inductance in A, B, C phase Δ t time interval respectively, as shown in formula (13)；
In formula, LmA(k)、LmB(k)、LmC(k) be respectively k-th of sampled point of A, B, C phase magnetizing inductance,The average value of magnetizing inductance respectively in A, B, C phase Δ t time interval, Respectively
For the variance of the magnetizing inductance in A, B, C phase Δ t time interval, n is the sampling number in Δ t time interval；
9) when the variance of the magnetizing inductance in any one phase Δ t time interval of A, B, C three-phase be greater than excitation surge current threshold value,
It is determined as that excitation surge current occurs, is latched three-stage percentage differential, is otherwise determined as the phase fault；
In formula, εA、εB、εCThe respectively excitation surge current threshold value of A, B, C phase.
(3) differential quick-break setting valve is calculated as shown in formula (15).
In formula, IAsdset、IBsdset、ICsdsetThe respectively differential quick-break setting valve of A, B, C phase；
I in formula (15)Aop、IBop、ICopIn any one phase when meeting operation condition, differential quick-break movement；
Step3, double-fed fan motor field are sent out transformer differential protection act and are differentiated, the action logic figure of three-phase differential protection is shown in attached drawing
Shown in 2.
(1) when any one phase meets operation condition in formula (15), protection act；
(2) any one phase meets operation condition and when formula (14) is unsatisfactory for operation condition in formula (2), protection act；
(3) when (1) and (2) is all unsatisfactory for, protection is failure to actuate.
The beneficial effects of the present invention are:
1, the present invention is according to the magnetizing inductance calculation formula of derivation, directly calculating magnetizing inductance numerical value, and seeks variance to it,
By variance calculated result compared with excitation surge current setting valve size, thus quickly and reliably differentiating transformer exciting surge and each
Kind internal fault.
2, the present invention combines excitation surge current locking criterion with three-stage percentage differential and differential quick-break, solves double-fed
Wind power plant was sent out when becoming internal fault because of second harmonic larger the problem of causing biased differential protection to be blocked, and can be effectively improved double
Present the reliability that wind power plant sends out transformer differential protection.
Detailed description of the invention
Fig. 1 is the three-phase transformer T shape equivalent circuit diagram of Y/ Δ wiring；
Fig. 2 is Three-Phase Transformer differential protection logic chart；
Rate of three phase flow differential action performance plot when Fig. 3 is the turn-to-turn short circuit of A phase；
Rate of three phase flow differential action performance plot when Fig. 4 is idle-loaded switching-on；
Fig. 5 is air-drop rate of three phase flow differential action performance plot when A phase turn-to-turn short circuit；
Three phase excitation inductance instantaneous value figure when Fig. 6 is the turn-to-turn short circuit of A phase；
Three phase excitation inductance instantaneous value figure when Fig. 7 is idle-loaded switching-on；
Fig. 8 is air-drop three phase excitation inductance instantaneous value figure when A phase turn-to-turn short circuit.
Specific embodiment
Embodiment 1: it is as shown in figures 1-8, a kind of to send out the differential protecting method become suitable for double-fed fan motor field, first with
Whether difference of phase currents Judging fault occurs；When not having failure, protection does not start, and when a failure occurs it, protection is opened
It is dynamic, three-stage percentage differential, excitation surge current locking, differential quick-break element movement logic judgment are carried out, when any one mutually differential speed
When disconnected action value is greater than differential quick-break setting valve, protection act；Any one phase three-stage percentage differential meets operation condition and one
When the variance for the magnetizing inductance fixed time in interval is less than setting valve, protection act.Otherwise, protection is failure to actuate.
Further, specific step is as follows for the differential protecting method that the double-fed fan motor field submitting becomes:
Step1, judge whether to break down according to difference of phase currents, as shown in formula (1).When any in A, B, C three-phase
One phase difference of phase currentsGreater than differential protection initiation value IcdWhen, it is judged as and breaks down,
Protection starting.Conversely, being judged as fault-free.
In formula,The difference of phase currents of respectively k-th sampled point A, B, C phase,The current value of respectively k-th sampled point A, B, C phase, Point
Not Wei previous cycle A, B, C phase current value,Respectively the first two period A,
B, the current value of C phase, N are the sampling number of a power frequency period, and k is k-th of sampled point, IcdFor differential protection initiation value, kk
For safety factor, kbFor load tap changer coefficient, kTAFor the error coefficient of current transformer TA, INFor high voltage side of transformer
Rated current.
Step2, judge failure occur after, differential protection starting, differential protection logic include three parts: percentage differential member
Part, shove blocking element and differential quick-break element.
(1) three-stage percentage differential
In formula, IopFor differential protection action current, Iop.minFor minimum working current, Ires.minFor the minimum restraint current,
IresFor stalling current, InFor Circuit Fault on Secondary Transformer rated current, k1For percentage differential coefficient.
A, the difference current of B, C phase and stalling current such as formula (3) and formula (4) are shown.
In formula, IAop、IBop、ICopRespectively transformer A, B, C phase difference current, IAres、IBres、ICresRespectively transformer
A, B, C phase stalling current, iA、iB、iCRespectively transformer primary side A, B, C phase current, ia、ib、icRespectively transformer secondary
Side A, B, C phase line current.
The difference current of A, B, C phase and stalling current in formula (3) and formula (4) are brought into respectively in formula (2), when in three-phase
When a phase difference current of anticipating meets operation condition, three-stage percentage differential element movement.
(2) it shoves locking
1) according to the three-phase transformer T shape equivalent circuit diagram (see attached drawing 1) of Y/ Δ wiring, three-phase loop equation is formed, such as
Shown in formula (5).
In formula (5), uA、uB、uCRespectively transformer Y shape side A, B, C phase voltage,Respectively transformer Δ side
A, B, C phase winding electric current, LA、LB、LCThe respectively side transformer Y A, B, C phase leakage inductance, RA、RB、RCRespectively transformer A, B, C phase
Resistance, RmA、RmB、RmCRespectively transformer A, B, C phase excitation resistance, LmA、LmB、LmCRespectively transformer A, B, C phase excitation electricity
Sense, ipFor the circulation in Δ winding, t is the time.
2)LA、RA、RmAIt is much smaller than LmA, LB、RB、RmBIt is much smaller than LmB, LC、RC、RmCIt is much smaller than LmC, therefore formula (5) can
Approximate representation is formula (6).
3) i is eliminated using formula (6)p, as shown in formula (7).
4) transformer Δ side A, B, C phase winding electric current in formula (7) is soughtAs shown in formula (8)-(9).
Solution formula (8) obtains Δ side A, B, C phase winding electric currentAs shown in formula (9).
5) formula (9) is brought into formula (7), as shown in formula (10).
6) it enablesIt is right
Formula (10) carries out discretization, as shown in formula (11).
In formula, uA(k)、uB(k)、uCIt (k) is respectively k-th of the side sampled point transformer Y A, B, C phase voltage, uA(k+1)、uB
(k+1)、uCIt (k+1) is respectively+1 side sampled point transformer Y A, B, C phase voltage of kth, Δ T is+1 sampled point of kth and kth
The time difference of a sampled point, i'abThe sum of the side Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcFor the side Y B, C two-phase
The sum of current difference and Δ side b, c biphase current difference, i 'caFor the sum of the side Y C, A biphase current difference and Δ side c, a biphase current difference,
i'abIt (k) is the sum of k-th of the side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k) is k-th of sampling
The sum of the side point Y B, C biphase current difference and Δ side b, c biphase current difference, i'ca(k) poor for k-th of the side sampled point Y C, A biphase current
The sum of with Δ side c, a biphase current difference, i'abIt (k+1) is+1 side sampled point Y A, B biphase current difference of kth and Δ side a, b two-phase
The sum of current difference, i 'bcIt (k+1) is the sum of+1 side sampled point Y B, C biphase current difference of kth and Δ side b, c biphase current difference,
i'caIt (k+1) is the sum of+1 side sampled point Y C, A biphase current difference of kth and Δ side c, a biphase current difference, i'abIt (k-1) is the
The sum of the k-1 side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k-1) is -1 side sampled point Y of kth
B, the sum of C biphase current difference and Δ side b, c biphase current difference, i'ca(k-1) poor for -1 side sampled point Y C, A biphase current of kth
The sum of with Δ side c, a biphase current difference, i'abIt (k+2) is+2 side sampled point Y A, B biphase current differences of kth and Δ side a, b two-phase
The sum of current difference, i'bcIt (k+2) is the sum of+2 side sampled point Y B, C biphase current differences of kth and Δ side b, c biphase current difference,
i′caIt (k+2) is respectively the sum of+2 side sampled point Y C, A biphase current differences of kth and Δ side c, a biphase current difference.
7) magnetizing inductance that formula (11) respectively obtain A, B, C phase is solved, as shown in formula (12).
In formula, Δi'abIt (k) is k-th of sampled point i'abDiscrete value, Δ i'abIt (k+1) is kth+1
Sampled point i'abDiscrete value, Δ i 'bcIt (k) is k-th of sampled point i 'bcDiscrete value, Δ i 'bcIt (k+1) is+1 sampled point i ' of kthbc
Discrete value, Δ i'caIt (k) is k-th of sampled point i 'caDiscrete value, i 'caIt (k+1) is+1 sampled point i ' of kthcaDiscrete value.
8) variance for solving the magnetizing inductance in A, B, C phase Δ t time interval respectively, as shown in formula (13).
In formula, LmA(k)、LmB(k)、LmC(k) be respectively k-th of sampled point of A, B, C phase magnetizing inductance,The average value of magnetizing inductance respectively in A, B, C phase Δ t time interval, Respectively
For the variance of the magnetizing inductance in A, B, C phase Δ t time interval, n is the sampling number in Δ t time interval.
9) when the variance of the magnetizing inductance in any one phase Δ t time interval of A, B, C three-phase be greater than excitation surge current threshold value,
It is determined as that excitation surge current occurs, is latched three-stage percentage differential, is otherwise determined as the phase fault.
In formula, εA、εB、εCThe respectively excitation surge current threshold value of A, B, C phase.
(3) differential quick-break setting valve is calculated as shown in formula (15).
In formula, IAsdset、IBsdset、ICsdsetThe respectively differential quick-break setting valve of A, B, C phase.
I in formula (15)Aop、IBop、ICopIn any one phase when meeting operation condition, differential quick-break movement.
Step3, double-fed fan motor field are sent out transformer differential protection act and are differentiated, the action logic figure of three-phase differential protection is shown in attached drawing
Shown in 2.
(1) when any one phase meets operation condition in formula (15), protection act.
(2) any one phase meets operation condition and when formula (14) is unsatisfactory for operation condition in formula (2), protection act.
(3) when (1) and (2) is all unsatisfactory for, protection is failure to actuate.
Concrete principle of the invention is as follows:
(1) double-fed fan motor unit short circuit current characteristic and the influence to change conventional differential protection is sent out.
When electric network fault cause set end voltage fall degree it is not deep when, the Crowbar circuit of double-fed blower is failure to actuate, blower
Low-voltage crossing is realized by the control strategy of itself.At this point, the short circuit current characteristic of double-fed blower and the control strategy of frequency converter
It is closely related.When electric network fault causes set end voltage seriously to fall, blower electromagnetic torque is reduced, and stable state vector controlled will increase
Rotor excitation current, aerator supervision rotor over-current, investment Crowbar protection circuit protect rotor-side converter, when stage
Between it is of short duration, generally continue 3~5ms, tranformer protection has little time to act.Hereafter double-fed blower is equivalent to common asynchronous generating
Machine is continued until that failure vanishes, Crowbar circuit are out of service.If three-phase shortcircuit, generator terminal fault current can during this period
Shown in approximate representation such as formula (1).
In formula, ikFor short circuit current, t is time, A1For fundamental frequency time coefficient, A2For attenuating dc component coefficient, A3To decline
Turn reducing speed component coefficient, size is related with the parameter of double-fed blower, the degree of Voltage Drop, and A3> A1, α is fault moment
The initial phase angle of stator voltage, ωsFor synchronous rotational speed, ωrFor rotor speed, TsFor the time attenuation constant of stator side, TrFor rotor
The time attenuation constant of side.
By formula (1) it is found that after investment Crowbar circuit, stator failure electric current ikBy stable state fundamental component, the transient state of decaying
DC component and damped alternating current component three parts are constituted.Wherein when Voltage Drop is deeper, at the beginning of damped alternating current component is failure
The main composition part of phase generator terminal electric current, frequency depend on current rotating speed.Double-fed type rotation speed of fan variation range is generally 0.7
~1.3p.u, therefore generator terminal output fault current is mainly made of rotor speed frequency component under different operating conditions, it will
Change within the scope of 35~65Hz.
The fundamental voltage amplitude and secondary harmonic amplitude of short circuit current are extracted in conventional differential protection using power frequency Fourier algorithm,
However the short circuit current frequency not necessarily power frequency of short dot wind farm side, therefore the current first harmonics that power frequency Fourier algorithm obtains
Amplitude and secondary harmonic amplitude can generating period fluctuations.
(2) double-fed fan motor field second_harmonic generation principle.
When short circuit occurs for power grid, due to the control action of stator magnetic linkage and rotor current transformer, stator and rotor winding generates respectively
Electric current containing forced component, transient DC component and transient state nature component, leads to DC bus-bar voltage and net side reactive power
It vibrating, net side is caused to carry out comprehensive regulation, voltage on line side and electric current will all generate 2 harmonics, by PWM modulation,
Rotor windings generate the harmonic current components of (1+s) f (s is the revolutional slip of double-fed blower, and f is the frequency of double-fed blower), finally
The harmonic current components of two frequencys multiplication are induced in stator winding.
(3) excitation surge current producing cause
1) transformer during no-load closing
Circuit Fault on Secondary Transformer is unloaded, first winding is accessed power supply, the referred to as idle-loaded switching-on of transformer, transformer is normally transported
When row, exciting current very little, the 2%~10% of general only rated current.But idle-loaded switching-on is to the moment of power grid, due to iron core
There are saturated phenomenon, exciting current may be sharply increased as tens times of normal current or even hundreds of times, what idle-loaded switching-on occurred
Transient current surge may cause system trip, so that transformer cannot smoothly put into power grid.
By taking single-phase transformer as an example, network voltage u1Change by sinusoidal rule, then the balance of voltage side of primary side when unloaded
Shown in journey such as formula (2).
In formula, u1For network voltage, U1For supply voltage virtual value,Voltage u when to close a floodgate1Initial phase angle, φ be and one
The total magnetic flux of secondary winding interlinkage, N1For the number of turns of first side winding, r1For the resistance of first winding, ω is angular frequency, i0For zero load
Electric current.
It is calculated to simplify, ignores transformer first winding resistance r1, and not considering iron core remanent magnetism, then formula (2) can letter
Turn to formula (3).
In t=0, under the primary condition of φ=0, shown in the solution such as formula (4) of formula (3).
In formula,φmMagnetic flux amplitude when for stable state,
φtFor the steady-state component of magnetic flux, φt' be magnetic flux transient state component.
Formula (4) shows voltage u when the size and combined floodgate of magnetic flux1Initial phase angleIt is related.Discuss separately below two kinds it is extreme
Situation.
1. initial phase angleWhen close a floodgate
Shown in transformer flux such as formula (5).
φ=φm sin(ωt) (5)
At this point, transient state component φt'=0 establishes stable state magnetic flux after combined floodgate immediately, without transient process, avoids unloaded conjunction
The generation of dash current when lock, that is to say, that transformer closes a floodgate the most advantageous in this case.
2. initial phase angleWhen close a floodgate
Shown in transformer flux such as formula (6).
φ=- φm cos(ωt)+φm (6)
At this point, in idle-loaded switching-on second half of the cycle (t=π/ω) moment, magnetic flux reaches maximum value, φmax=2 φm, it is positive
2 times of normal magnetic flux, this 2 times magnetic flux will make iron core be in serious supersaturation, sharply increase so as to cause exciting current,
It can reach dozens or even hundreds of times of normal exciting current, 5~8 times of rated current.Iron core degree of saturation is higher, switching current
Also bigger.
2) Circuit Fault on Secondary Transformer suddenly-applied short circuit
Single-phase transformer suddenly-applied short circuit, shown in circuit equation such as formula (7).
In formula, β is voltage u when suddenly-applied short circuit occurs1Initial phase angle, LkFor the leakage inductance of transformer, rkFor the short circuit of transformer
Resistance.
Formula (7) is constant coefficient differential equation of first order, ignores no-load current and load current, that is, when thinking t=0, ik=0,
And ω Lk> > rk, shown in non trivial solution such as formula (8).
In formula, IkFor steady-state component current effective value,TkFor transient state component decaying time constant,xkFor the short-circuit impedance of transformer, i'kFor electric current steady-state component, i "kFor the transient state component of electric current
The size of sudden short-circuit current is related with the initial phase angle of transcient short circuit time supply voltage occurs it can be seen from formula (9),
Two kinds of extreme cases are discussed separately below.
1. suddenly-applied short circuit occurs to go out phase angle in voltageWhen, shown in short circuit current such as formula (9).
At this point, transient state component i "k=0, suddenly-applied short circuit one enters stable state, and short-circuit current value is minimum.
2. suddenly-applied short circuit occurs in voltage initial angle β=0 °, shown in short circuit current such as formula (10).
Its maximum value occurs to carry it into formula (10) in short-circuit second half of the cycle moment, i.e. t=π/ω, can find out maximum
Short circuit current, as shown in formula (11).
In formula, ikmaxFor maximum short circuit current, kyFor a part in calculation of short-circuit current formula,It is prominent
The maximum value of right short circuit current and the ratio of steady-state shortcircuit current maximum value, size are decided by the parameter r of transformerkAnd xk, right
In Transformer, there is ky=1.2~1.4, for high-power transformer, there is ky=1.5~1.8.
Suddenly-applied short circuit can cause transformer to generate very big dash current, influence of this overcurrent to transformer mainly by
Two aspects, first is that electromagnetic force is generated, second is that transformer is made to generate heat.
(4) magnetizing inductance numerical characteristics
When transformer core saturation, the slope of flux curve is smaller, therefore magnetizing inductance numerical value is smaller, seriously encourages when saturation
Magnetoelectricity inductance value is close to air core inductor value；When excitation surge current occurs, transformer core periodically enters saturation region and exits saturation
Area causes magnetizing inductance periodicity change dramatically value larger, i.e., magnetizing inductance fluctuates between the larger value and smaller value；In inside
When failure, due to the presence of fault branch or failure ring, the numerical value of magnetizing inductance is smaller, and does not have fluctuation.According to excitation
The size and fluctuation of inductance value identify the irregular operatings state such as internal fault and excitation surge current.
It elaborates below with reference to specific example to the present invention.
For this example by taking the wind power plant of 20 double-fed blowers as an example, the parameter of wind power plant sends out the parameter become and differential protection
Parameter has divided following three kinds of situations to implement as shown in table 1, table 2 and table 3.
(1) it is sent out in 1.05s-1.2s and becomes the generation A phase turn-to-turn short circuit of Δ side
(2) it is sent out in 1.05s and becomes idle-loaded switching-on
(3) become Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
The basic parameter of 1 double-fed fan motor field of table
Send out variable element in 2 double-fed fan motor field of table
3 differential protection parameter of table
Specifically carry out as follows:
(1) judge whether to break down according to difference of phase currents, as shown in formula (1).When phase any one in A, B, C three-phase
Difference of phase currentsGreater than differential protection initiation value IcdWhen, it is judged as and breaks down, protects
Starting.Conversely, being judged as fault-free.
In formula,The difference of phase currents of respectively k-th sampled point A, B, C phase,The current value of respectively k-th sampled point A, B, C phase, Point
Not Wei previous cycle A, B, C phase current value,Respectively the first two period A,
B, the current value of C phase, N are the sampling number of a power frequency period, and k is k-th of sampled point, IcdFor differential protection initiation value, kk
For safety factor, kbFor load tap changer coefficient, kTAFor the error coefficient of current transformer TA, INFor high voltage side of transformer
Rated current.
Three kinds of situation fault distinguishing results are as follows.
1) it is sent out in 1.05s-1.2s and becomes the generation A phase turn-to-turn short circuit of Δ side
The difference of phase currents of B phase is less than differential starting protection value Icd, the difference of phase currents of A phase and C phase is greater than differential
Starting protection value Icd, it is determined as breaking down, protection starting.
2) it is sent out in 1.05s and becomes idle-loaded switching-on
A, the difference of phase currents of B phase is less than differential starting protection value Icd, the difference of phase currents of C phase is greater than differential starting
Protection value Icd, it is determined as breaking down, protection starting.
3) become Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
The difference of phase currents of B phase is less than differential starting protection value Icd, the difference of phase currents of A phase and C phase is greater than differential
Starting protection value Icd, it is determined as breaking down, protection starting.
(2) after judging that failure occurs, differential protection starting, differential protection logic includes three parts: percentage differential element,
Shove blocking element and differential quick-break element.
1) three-stage percentage differential
In formula, IopFor differential protection action current, Iop.minFor minimum working current, Ires.minFor the minimum restraint current,
IresFor stalling current, InFor Circuit Fault on Secondary Transformer rated current, k1For percentage differential coefficient.
A, the difference current of B, C phase and stalling current such as formula (3) and formula (4) are shown.
In formula, IAop、IBop、ICopRespectively transformer A, B, C phase difference current, IAres、IBres、ICresRespectively transformer
A, B, C phase stalling current, iA、iB、iCRespectively transformer primary side A, B, C phase current, ia、ib、icRespectively transformer secondary
Side A, B, C phase line current.
The difference current of A, B, C phase and stalling current in formula (3) and formula (4) are brought into respectively in formula (2), when in three-phase
When a phase difference current of anticipating meets operation condition, three-stage percentage differential element movement.
Three kinds of situation three-stage percentage differential movements differentiate that result is as follows.
The turn-to-turn short circuit of A phase occurs 1. sending out in 1.05s-1.2s and becoming Δ side
From attached drawing 3 it is found that being greater than braking amount in the actuating quantity of failure A early period phase and C phase, between 1.084s-1.109s,
A phase actuating quantity is less than braking amount, and between 1.109s-1.2s, A phase actuating quantity is greater than braking amount；Between 1.095s-1.11s,
C phase actuating quantity is less than braking amount, and between 1.11s-1.2s, C phase actuating quantity is greater than braking amount；The B phase actuating quantity during failure
Both less than braking amount.
2. being sent out in 1.05s and becoming idle-loaded switching-on
From attached drawing 4 it is found that during 1.05s-1.3s, the actuating quantity of B phase and C phase is all larger than braking amount, and A phase actuating quantity is all
Less than braking amount.
3. becoming Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
From attached drawing 5 it is found that during failure, the actuating quantity of A phase and C phase is all larger than braking amount, and B phase actuating quantity is both less than made
Momentum.
2) it shoves locking
1. forming three-phase loop equation, such as according to the three-phase transformer T shape equivalent circuit diagram (see attached drawing 1) of Y/ Δ wiring
Shown in formula (5).
In formula (5), uA、uB、uCRespectively transformer Y shape side A, B, C phase voltage,Respectively transformer Δ side
A, B, C phase winding electric current, LA、LB、LCThe respectively side transformer Y A, B, C phase leakage inductance, RA、RB、RCRespectively transformer A, B, C phase
Resistance, RmA、RmB、RmCRespectively transformer A, B, C phase excitation resistance, LmA、LmB、LmCRespectively transformer A, B, C phase excitation electricity
Sense, ipFor the circulation in Δ winding, t is the time.
②LA、RA、RmAIt is much smaller than LmA, LB、RB、RmBIt is much smaller than LmB, LC、RC、RmCIt is much smaller than LmC, therefore formula (5) can
Approximate representation is formula (6).
3. eliminating i using formula (6)p, as shown in formula (7).
4. seeking transformer Δ side A, B, C phase winding electric current in formula (7)As shown in formula (8)-(9).
Solution formula (8) obtains Δ side A, B, C phase winding electric currentAs shown in formula (9).
5. formula (9) is brought into formula (7), as shown in formula (10).
6. enablingIt is right
Formula (10) carries out shown in discretization such as formula (11).
In formula, uA(k)、uB(k)、uCIt (k) is respectively k-th of the side sampled point transformer Y A, B, C phase voltage, uA(k+1)、uB
(k+1)、uCIt (k+1) is respectively+1 side sampled point transformer Y A, B, C phase voltage of kth, Δ T is+1 sampled point of kth and kth
The time difference of a sampled point, i'abThe sum of the side Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcFor the side Y B, C two-phase
The sum of current difference and Δ side b, c biphase current difference, i 'caFor the sum of the side Y C, A biphase current difference and Δ side c, a biphase current difference,
i'abIt (k) is the sum of k-th of the side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k) is k-th of sampling
The sum of the side point Y B, C biphase current difference and Δ side b, c biphase current difference, i'ca(k) poor for k-th of the side sampled point Y C, A biphase current
The sum of with Δ side c, a biphase current difference, i'abIt (k+1) is+1 side sampled point Y A, B biphase current difference of kth and Δ side a, b two-phase
The sum of current difference, ib'cIt (k+1) is the sum of+1 side sampled point Y B, C biphase current difference of kth and Δ side b, c biphase current difference,
i'caIt (k+1) is the sum of+1 side sampled point Y C, A biphase current difference of kth and Δ side c, a biphase current difference, i'abIt (k-1) is the
The sum of the k-1 side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k-1) is -1 side sampled point Y of kth
B, the sum of C biphase current difference and Δ side b, c biphase current difference, i'ca(k-1) poor for -1 side sampled point Y C, A biphase current of kth
The sum of with Δ side c, a biphase current difference, i'abIt (k+2) is+2 side sampled point Y A, B biphase current differences of kth and Δ side a, b two-phase
The sum of current difference, i'bcIt (k+2) is the sum of+2 side sampled point Y B, C biphase current differences of kth and Δ side b, c biphase current difference,
i′caIt (k+2) is respectively the sum of+2 side sampled point Y C, A biphase current differences of kth and Δ side c, a biphase current difference.
7. solving formula (11) to respectively obtain shown in the magnetizing inductance such as formula (12) of A, B, C phase.
In formula, Δi'abIt (k) is k-th of sampled point i'abDiscrete value, Δ i'abIt (k+1) is kth+1
Sampled point i'abDiscrete value, Δ i 'bcIt (k) is k-th of sampled point i 'bcDiscrete value, Δ i 'bcIt (k+1) is+1 sampled point i ' of kthbc
Discrete value, Δ i'caIt (k) is k-th of sampled point i 'caDiscrete value, i'caIt (k+1) is+1 sampled point i ' of kthcaDiscrete value.
8. the variance of the magnetizing inductance in the Δ t time interval of A, B, C phase is solved respectively, as shown in formula (13).
In formula, LmA(k)、LmB(k)、LmC(k) be respectively k-th of sampled point of A, B, C phase magnetizing inductance,The average value of magnetizing inductance respectively in one period of A, B, C phase,Respectively A,
B, the variance of magnetizing inductance in one period of C phase, n are the sampling number in Δ t time interval.
9. the variance when the magnetizing inductance in any one phase Δ t time interval of A, B, C three-phase is greater than excitation surge current threshold value,
It is determined as that excitation surge current occurs, is latched three-stage percentage differential, is otherwise determined as the phase fault.
In formula, εA、εB、εCThe respectively excitation surge current threshold value of A, B, C phase.
In the step Step2 excitation surge current locking, Δ t=0.02s is divided between setting time, according to analysis of simulation experiment
Method determine εA=300, εB=300, εC=300
Three kinds of situation excitation surge current block actions differentiate that result is as follows.
A) it is sent out in 1.05s-1.2s and becomes the generation A phase turn-to-turn short circuit of Δ side
A, the excitation in one period of B, C three-phase
Inductance variance yields is both less than excitation surge current threshold value, and excitation surge current blocking element is failure to actuate, from failure generation A in a period,
B, the results are shown in attached figure 6 for C phase magnetizing inductance instantaneous value.
B) it is sent out in 1.05s and becomes idle-loaded switching-on
A, the excitation in one period of B phase
Inductance variance yields is both less than excitation surge current threshold value, and the magnetizing inductance variance yields in one period of C phase is greater than excitation surge current threshold value,
The movement of excitation surge current blocking element, the results are shown in attached figure 7 for A, B, C phase magnetizing inductance instantaneous value in a period from failure generation.
C) become Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
A, in one period of B, C three-phase
Magnetizing inductance variance yields is both less than excitation surge current threshold value, and excitation surge current blocking element is failure to actuate, a period from failure generation
Interior A, B, C phase magnetizing inductance instantaneous value the results are shown in attached figure 8
(3) differential quick-break setting valve is calculated as shown in formula (15).
In formula, IAsdset、IBsdset、ICsdsetThe respectively differential quick-break setting valve of A, B, C phase.
I in formula (15)Aop、IBop、ICopIn any one phase when meeting operation condition, then differential quick-break movement.
The differential quick-break movement of three kinds of situations differentiates that result is as follows.
1) it is sent out in 1.05s-1.2s and becomes the generation A phase turn-to-turn short circuit of Δ side
A, the differential quick-break action value of B, C three-phase is both less than differential quick-break setting valve, and differential quick-break element is failure to actuate.
2) it is sent out in 1.05s and becomes idle-loaded switching-on
A, the differential quick-break action value of B, C three-phase is both less than differential quick-break setting valve, and differential quick-break element is failure to actuate.
3) become Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
A, the differential quick-break action value of B, C three-phase is both less than differential quick-break setting valve, and differential quick-break element is failure to actuate.
Step3, double-fed fan motor field are sent out transformer differential protection act and are differentiated, the action logic figure of three-phase differential protection is shown in attached drawing
Shown in 2.
(1) when any one phase meets operation condition in formula (15), protection act.
(2) any one phase meets operation condition and when formula (14) is unsatisfactory for operation condition in formula (2), protection act.
(3) when (1) and (2) is all unsatisfactory for, protection is failure to actuate.
The differential quick-break movement of three kinds of situations differentiates that result is as follows.
1) it is sent out in 1.05s-1.2s and becomes the generation A phase turn-to-turn short circuit of Δ side
Three-phase is all unsatisfactory for operation condition in formula (15), and A, C two-phase meet operation condition in formula (2) and formula (14) is unsatisfactory for
The differential protection movement become is sent out in operation condition, double-fed fan motor field.
2) it is sent out in 1.05s and becomes idle-loaded switching-on
Three-phase is all unsatisfactory for operation condition in formula (15), and B, C two-phase meet operation condition in formula (2) but formula (14) satisfaction is dynamic
Make condition, double-fed fan motor field sends out the differential protection become and is failure to actuate.
3) become Δ side A phase turn-to-turn short circuit in submitting in 1.05s idle-loaded switching-on
Three-phase is all unsatisfactory for operation condition in formula (15), and A, C two-phase meet operation condition in formula (2) and formula (14) is unsatisfactory for
The differential protection movement become is sent out in operation condition, double-fed fan motor field.
Above in conjunction with attached drawing, the embodiment of the present invention is explained in detail, but the present invention is not limited to above-mentioned
Embodiment within the knowledge of a person skilled in the art can also be before not departing from present inventive concept
Put that various changes can be made.
Claims (2)
1. a kind of send out the differential protecting method become suitable for double-fed fan motor field, it is characterised in that: be mutated first with phase current
Whether amount Judging fault occurs；When not having failure, protection does not start, and when a failure occurs it, protection starting carries out three sections
Formula percentage differential, excitation surge current locking, differential quick-break element movement logic judgment, when any one mutually differential quick-break action value is greater than
When differential quick-break setting valve, protection act；Any one phase three-stage percentage differential meets in operation condition and certain time interval
When the variance of magnetizing inductance is less than setting valve, protection act, otherwise, protection are failure to actuate.
2. the differential protecting method become is sent out in double-fed fan motor field according to claim 1, it is characterised in that: the double-fed wind
Specific step is as follows for the differential protecting method that electric field submitting becomes:
Step1, judge whether to break down according to difference of phase currents, as shown in formula (1), when phase any one in A, B, C three-phase
Difference of phase currentsGreater than differential protection initiation value IcdWhen, it is judged as and breaks down, protects
Starting, conversely, being judged as fault-free；
In formula,The difference of phase currents of respectively k-th sampled point A, B, C phase,The current value of respectively k-th sampled point A, B, C phase,
The respectively current value of previous cycle A, B, C phase,Respectively the first two period
A, the current value of B, C phase, N are the sampling number of a power frequency period, and k is k-th of sampled point, IcdFor differential protection initiation value,
kkFor safety factor, kbFor load tap changer coefficient, kTAFor the error coefficient of current transformer TA, INFor high voltage side of transformer
Rated current；
Step2, judge failure occur after, differential protection starting, differential protection logic include three parts: percentage differential element,
Shove blocking element and differential quick-break element；
(1) three-stage percentage differential
In formula, IopFor differential protection action current, Iop.minFor minimum working current, Ires.minFor the minimum restraint current, IresFor
Stalling current, InFor Circuit Fault on Secondary Transformer rated current, k1For percentage differential coefficient；
A, the difference current of B, C phase and stalling current such as formula (3) and formula (4) are shown；
In formula, IAop、IBop、ICopRespectively transformer A, B, C phase difference current, IAres、IBres、ICresRespectively transformer A, B,
C phase stalling current, iA、iB、iCRespectively transformer primary side A, B, C phase current, ia、ib、icRespectively Circuit Fault on Secondary Transformer A,
B, C phase line current；
The difference current of A, B, C phase and stalling current in formula (3) and formula (4) are brought into respectively in formula (2), when any one in three-phase
When phase difference current meets operation condition, three-stage percentage differential element movement；
(2) it shoves locking
1) according to the three-phase transformer T shape equivalent circuit diagram of Y/ Δ wiring, three-phase loop equation is formed, as shown in formula (5)；
In formula (5), uA、uB、uCRespectively transformer Y shape side A, B, C phase voltage,Respectively transformer Δ side A, B, C
Phase winding electric current, LA、LB、LCThe respectively side transformer Y A, B, C phase leakage inductance, RA、RB、RCRespectively transformer A, B, C phase resistance,
RmA、RmB、RmCRespectively transformer A, B, C phase excitation resistance, LmA、LmB、LmCRespectively transformer A, B, C phase magnetizing inductance, ip
For the circulation in Δ winding, t is the time；
2)LA、RA、RmAIt is much smaller than LmA, LB、RB、RmBIt is much smaller than LmB, LC、RC、RmCIt is much smaller than LmC, therefore formula (5) can be approximate
It is expressed as formula (6)；
3) i is eliminated using formula (6)p, as shown in formula (7)；
4) transformer Δ side A, B, C phase winding electric current in formula (7) is soughtAs shown in formula (8)-(9)；
Solution formula (8) obtains Δ side A, B, C phase winding electric currentAs shown in formula (9)；
5) formula (9) is brought into formula (7), as shown in formula (10)；
6) it enablesTo formula
(10) discretization is carried out, as shown in formula (11)；
In formula, uA(k)、uB(k)、uCIt (k) is respectively k-th of the side sampled point transformer Y A, B, C phase voltage, uA(k+1)、uB(k+
1)、uCIt (k+1) is respectively+1 side sampled point transformer Y A, B, C phase voltage of kth, Δ T is that+1 sampled point of kth is adopted with k-th
The time difference of sampling point, i 'abThe sum of the side Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcFor the side Y B, C biphase current
The sum of difference and Δ side b, c biphase current difference, i 'caFor the sum of the side Y C, A biphase current difference and Δ side c, a biphase current difference, i 'ab
It (k) is the sum of k-th of the side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k) is k-th of sampled point Y
The sum of side B, C biphase current difference and Δ side b, c biphase current difference, i 'ca(k) for k-th side sampled point Y C, A biphase current difference with
The sum of Δ side c, a biphase current difference, i 'abIt (k+1) is+1 side sampled point Y A, B biphase current difference of kth and Δ side a, b two-phase electricity
Flow the sum of difference, i 'bcIt (k+1) is the sum of+1 side sampled point Y B, C biphase current difference of kth and Δ side b, c biphase current difference, i 'ca
It (k+1) is the sum of+1 side sampled point Y C, A biphase current difference of kth and Δ side c, a biphase current difference, i 'abIt (k-1) is kth -1
The sum of the side sampled point Y A, B biphase current difference and Δ side a, b biphase current difference, i 'bcIt (k-1) is -1 side sampled point Y B, C two of kth
The sum of phase differential current and Δ side b, c biphase current difference, i 'caIt (k-1) is -1 side sampled point Y C, A biphase current difference of kth and Δ side
C, the sum of a biphase current difference, i 'ab(k+2) poor for+2 side sampled point Y A, B biphase current differences of kth and Δ side a, b biphase current
The sum of, i 'bcIt (k+2) is the sum of+2 side sampled point Y B, C biphase current differences of kth and Δ side b, c biphase current difference, i 'ca(k+2)
Respectively the sum of+2 side sampled point Y C, A biphase current differences of kth and Δ side c, a biphase current difference；
7) magnetizing inductance that formula (11) respectively obtain A, B, C phase is solved, as shown in formula (12)；
In formula, Δi′abIt (k) is k-th of sampled point i 'abDiscrete value, Δ i 'abIt (k+1) is kth+1
Sampled point i 'abDiscrete value, Δ i 'bcIt (k) is k-th of sampled point i 'bcDiscrete value, Δ i 'bcIt (k+1) is+1 sampled point i ' of kthbc
Discrete value, Δ i 'caIt (k) is k-th of sampled point i 'caDiscrete value, i 'caIt (k+1) is+1 sampled point i ' of kthcaDiscrete value；
8) variance for solving the magnetizing inductance in A, B, C phase Δ t time interval respectively, as shown in formula (13)；
In formula, LmA(k)、LmB(k)、LmC(k) be respectively k-th of sampled point of A, B, C phase magnetizing inductance,Point
Not Wei magnetizing inductance in A, B, C phase Δ t time interval average value, When respectively A, B, C phase Δ t
Between interval in magnetizing inductance variance, n be Δ t time interval in sampling number；
9) when the variance of the magnetizing inductance in any one phase Δ t time interval of A, B, C three-phase is greater than excitation surge current threshold value, differentiation
For excitation surge current occurs, it is latched three-stage percentage differential, is otherwise determined as the phase fault；
In formula, εA、εB、εCThe respectively excitation surge current threshold value of A, B, C phase；
(3) differential quick-break setting valve is calculated as shown in formula (15)；
In formula, IAsdset、IBsdset、ICsdsetThe respectively differential quick-break setting valve of A, B, C phase；
I in formula (15)Aop、IBop、ICopIn any one phase when meeting operation condition, differential quick-break movement；
Step3, double-fed fan motor field are sent out transformer differential protection act and are differentiated；
(1) when any one phase meets operation condition in formula (15), protection act；
(2) any one phase meets operation condition and when formula (14) is unsatisfactory for operation condition in formula (2), protection act；
(3) when (1) and (2) is all unsatisfactory for, protection is failure to actuate.
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