CN104899677A

CN104899677A - Quick estimation method for load maximum reactive power command of 500kv radial power supply network in disruption recovery

Info

Publication number: CN104899677A
Application number: CN201510226883.2A
Authority: CN
Inventors: 毛晓明; 吴杰康; 洪俊杰; 郭壮志
Original assignee: Guangdong University of Technology
Current assignee: Guangdong University of Technology
Priority date: 2015-05-06
Filing date: 2015-05-06
Publication date: 2015-09-09
Anticipated expiration: 2035-05-06
Also published as: CN104899677B

Abstract

The invention discloses a quick estimation method for a load maximum reactive power command of a 500kv radial power supply network in disruption recovery. The method comprises the following steps: 1) calculating operating data under a steady-state condition: 1.1, calculating the equivalent impedance of a composite load of a main transformer; 1.2, calculating the equivalent impedance per unit value of an induction motor; 1.3, calculating the equivalent impedance of a constant impedance load; 2) calculating relevant parameters of an equivalent circuit of the 500kV transformer power supply network; 3) calculating relevant parameters at the fault clearing moment; 4) estimating 10kV bus voltages of 200kV and 110kV transformers at the fault clearing moment; and 5) estimating the maximum dynamic reactive power command of the composite load of the 500kV main transformer in recovery after the main network fault. The method needs a few basic data for calculation, and is quick, time-saving and labor-saving.

Description

Method for quickly estimating load maximum reactive power demand of 500kV radial power supply network in disturbance recovery

Technical Field

The invention relates to a method for quickly estimating the maximum reactive power demand of a load of a 500kV radial power supply network in disturbance recovery, in particular to a method for quickly estimating the maximum reactive power demand of a comprehensive load (approximately considered to occur at the moment of fault removal) of a 500kV substation radial power supply network after a main network fault.

Background

In the analysis and calculation of the power system, power users which are difficult to win in a wide area covered by a power grid are often combined into a small number of comprehensive loads, and the comprehensive loads are tapped on buses of different voltage levels of a transformer substation. In power system voltage stability studies, the combined load is often expressed as a combination of constant impedance and induction motor of different proportions.

When the high-voltage side of the main transformer has a short-circuit fault, the relay protection device of the power system can rapidly remove the fault. In the process of recovery after a fault, the comprehensive load can generate obvious dynamic reactive power demand, and the size of the dynamic reactive power demand depends on the type of the fault, the duration time of the fault, the load characteristic and the voltage recovery of a system; the system voltage recovery is affected by the alternating current system strength and the load dynamic reactive power requirement, and the alternating current system strength and the load dynamic reactive power requirement are coupled with each other. For the dynamic reactive power demand of the comprehensive load, the calculation can be realized more accurately only by means of calculation and analysis software of the power system, and the method is time-consuming and labor-consuming.

Disclosure of Invention

The invention aims to solve the technical problem of providing a method for quickly estimating the maximum reactive power demand of the load of a 500kV radial power supply network in disturbance recovery, aiming at the fault type of the most serious three-phase metallic grounding short circuit of a power system, and having the advantages of less basic data, high speed, time saving and labor saving.

The technical scheme adopted by the invention is as follows:

a method for quickly estimating the maximum reactive power demand of a 500kV radial power supply network load in disturbance recovery comprises the following basic data:

1) the method comprises the following steps that (1) the number n of main transformers of a 500kV transformer substation and the short-circuit current level Id of a high-voltage bus are in a unit kA;

2) active power proportion K of induction motor in transformer substation main transformer high-voltage side active power P, unit MW and comprehensive load during normal operation_P；

3) Inertia time constant T of induction motor_jLoad factor K_LAnd equivalent electric road single parameter: stator resistance R_sStator reactance X_sExcitation reactance X_MRotor reactance X_RRotor resistance R_R；

4) The fault duration Δ t;

the method requires the use of the following basic formula:

1) basic formula of induction motor

Referring to fig. 1, the induction motor has the equivalent: stator resistors R connected in series in sequence_s、Stator reactance X_s、Rotor reactance X_RAnd a variable resistance R ═ R_RS, in rotor reactance X_RAnd a variable resistance R ═ R_RThe two ends of/s are provided with excitation reactance X_M；

The equivalent impedance Z of the induction motor is calculated as formula (1):

Z = R_{S} + j X_{S} + \frac{(j X_{M}) (R + j X_{R})}{R + j (X_{M} + X_{R})} - - - (1);

the defining variables are as in formula (2):

\{\begin{matrix} X_{RM} = X_{R} + X_{M} \\ X_{SM} = X_{S} + X_{M} \\ X_{P} = X_{S} X_{R} + X_{S} X_{M} + X_{M} X_{R} \end{matrix} - - - (2);

the equivalent admittance Y of the induction motor is calculated as formula (3):

\{\begin{matrix} Y = \frac{1}{Z} = \frac{R + j X_{RM}}{R_{S} R - X_{P} + j (R_{S} X_{RM} + R X_{SM})} \\ = \frac{(R_{S} R^{2} + X_{M}^{2} R + R_{S} X_{RM}^{2}) - j (X_{SM} R^{2} + X_{P} X_{RM})}{A R^{2} + BR + C} \end{matrix} - - - (3);

wherein,

\{\begin{matrix} A = R_{S}^{2} + X_{SM}^{2} \\ B = 2 R_{S} X_{M}^{2} \\ C = X_{P}^{2} + R_{S}^{2} X_{RM}^{2} \end{matrix} - - - (4);

when the operating voltage of the induction motor is V, the consumed power P + jQ is calculated as the formula (5):

\{\begin{matrix} P = REAL (V^{2} \overset{*}{Y}) = \frac{V^{2} (R_{S} R^{2} + X_{M}^{2} R + R_{S} X_{RM}^{2})}{A R^{2} + BR + C} \\ Q = IMAG (V^{2} \overset{*}{Y}) = \frac{V^{2} (X_{SM} R^{2} + X_{P} X_{RM})}{A R^{2} + BR + C} \end{matrix} - - - (5);

2) equivalent impedance calculation formula of electrical element

If the operating voltage of the electrical element is V and the consumed power is P + jQ, the equivalent impedance Z of the electrical element is r + jx, and the equation (6) is calculated:

\{\begin{matrix} r = \frac{P}{P^{2} + Q^{2}} V^{2} \\ x = \frac{Q}{P^{2} + Q^{2}} V^{2} \end{matrix} - - - (6);

the method comprises the following steps:

1) running data calculation under steady state conditions

1.1 calculation of the equivalent impedance of the integrated load of the main transformer

According to the active power P on the high-voltage side of the main transformer of the transformer substation in normal operation in basic data, the unit MW is considered, P is the active power consumed by the comprehensive load on the 10kV side of the main transformer, the power factor is 0.95, and the reactive power Q (cos) consumed by the comprehensive load on the 10kV side of the main transformer with the parallel capacitance is obtained^-1(0.95)). And (3) obtaining the comprehensive load equivalent impedance Z on the 10kV side of the main transformer according to the formula (6) by taking the comprehensive load operating voltage V as 1p.u_Σ0；

1.2 Induction Motor equivalent impedance per Unit value calculation

Taking the terminal voltage V of an induction motor as 1p.u., and taking the active power P as K_Lp.u., according to formulae (2) and (4), and let:

\{\begin{matrix} A^{'} = \frac{AP}{V^{2}} - R_{s} \\ B^{'} = \frac{BP}{V^{2}} - X_{M}^{2} \\ C^{'} = \frac{CP}{V^{2}} - R_{s} X_{RM}^{2} \end{matrix} - - - (7);

obtaining equivalent resistance of the rotor of the induction motor in normal operation:

<math> <mrow> <msub> <mi>R</mi> <mn>0</mn> </msub> <mo>=</mo> <mfrac> <mrow> <mo>-</mo> <msup> <mi>B</mi> <mo>'</mo> </msup> <mo>&PlusMinus;</mo> <msqrt> <msup> <mi>B</mi> <mrow> <mo>'</mo> <mn>2</mn> </mrow> </msup> <mo>-</mo> <mn>4</mn> <msup> <mi>A</mi> <mo>'</mo> </msup> <msup> <mi>C</mi> <mo>'</mo> </msup> </msqrt> </mrow> <mrow> <mn>2</mn> <msup> <mi>A</mi> <mo>'</mo> </msup> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>

initial slip of induction motor:

s_{0} = \frac{R_{R}}{R_{0}} - - - (9);

calculating the equivalent impedance per unit value Z of the induction motor by the formula (1)_M*；

1.3 constant impedance load equivalent impedance Z_Z

According to the active power P of the high-voltage side of the transformer substation main transformer in normal operation, unit MW and the active power ratio K of the induction motor in the comprehensive load given in the basic data_PKnown as K_PP is the active power consumed by the motor, according to Z_M*Corresponding power factor angle, to obtain the reactive power consumed by the motor

Slave masterSubtracting the power consumed by the induction motor from the total load power to obtain the power consumed by the constant impedance load, and then obtaining the equivalent impedance Z of the constant impedance load by using the step (6)_Z；

2) Calculation of equivalent circuit related parameters of 500kV transformer substation power supply network

A common form of a 500kV substation supply network is shown in figure 2 a: the 500kV transformer substation supplies power to the 220kV transformer substation, and a comprehensive load is connected to the 10kV side of the 220kV transformer substation; the 220kV transformer substation supplies power to the 110kV transformer substation, and a comprehensive load is connected to the 10kV side of the 110kV transformer substation; FIG. 2a is equivalent to FIG. 2 b: the main network power supply is connected with the power supply through a transformation ratio of 1: e equivalent reactance Xs of an ideal transformer series system, and equivalent impedance X of a power transmission loop to a comprehensive load Z_Σ0The form of the power supply;

the system equivalent reactance Xs is estimated according to equation (10):

estimating the electrical distance between the comprehensive load and the high-voltage side of the 500kV main transformer according to the X-140 omega;

estimating the power supply return road sign unitary transformation ratio E according to the condition that the comprehensive load terminal voltage is 1p.u. during normal operation as shown in the formula (11):

3) calculation of relevant parameters at fault removal time

Induction motor terminal voltage during fault is considered to be zero, induction motor electromagnetic torque T_eInduction motor load torque T of 0_load＝K_LAnd delta t is the fault duration, and the induction motor rotating speed change delta omega at the fault removal moment is obtained according to the step (12):

the induction motor slip at the end of the fault is obtained according to equation (13):

s＝s₀-Δω (13)；

induction motor rotor resistance at the time of failure completion:

R = \frac{R_{R}}{s} - - - (14);

reference value of power of induction motor:

calculating equivalent impedances of various induction motor loads in the comprehensive loads at the moment of fault removal according to the formula (1); integrated load equivalent impedance Z_ΣCalculating the parallel value of the induction motor and the constant impedance load according to the formula (16):

Z_∑＝∏(Z) (16)；

II represents parallel connection;

4) estimation of bus voltage on 10kV side of 220kV and 110kV transformer substations at fault removal time

5) Estimation of maximum dynamic reactive power demand of 500kV main transformer comprehensive load during main network fault

Will U⁽¹⁰⁾And (5) substituting the reactive calculation formula in the formula to obtain the maximum dynamic reactive value of various induction motors in the 500kV main transformer comprehensive load, and summing to obtain the maximum dynamic reactive requirement of the 500kV main transformer comprehensive load when the main network fails.

Remarking: 1) about 20% of the comprehensive loads are supplied by the low-voltage side of a 220kV transformer substation, the part of the loads are closer to the high-voltage side of a 500kV main transformer, and all the loads in the algorithm are supplied according to the low-voltage side of a 110kV transformer substation, so that the estimated maximum dynamic reactive power demand of the comprehensive loads is lower; 2) at the moment of fault removal, the voltage of a load bus is low, reactive power taken by equivalent impedance of a load and a parallel capacitor except an induction motor is correspondingly reduced, and the influence of the reactive power taken by the equivalent impedance of the load and the parallel capacitor is ignored in an algorithm, so that the maximum dynamic reactive power demand of the estimated comprehensive load is high; the two effects substantially cancel.

Has the advantages that: the maximum reactive power demand of the load of the 500kV radial power supply network in disturbance recovery is generally solved by adopting a simulation means, time and labor are wasted, and no relevant analytical calculation method is reported at present. The invention has the advantages of less basic data required by calculation, high speed, time saving and labor saving.

Drawings

FIG. 1 is an equivalent circuit diagram of an induction motor;

figure 2a is a diagram of a typical 500kV substation supply network;

FIG. 2b is an equivalent circuit of a typical 500kV substation power supply network diagram;

FIG. 3 is a system wiring and tidal flow diagram of an embodiment;

FIG. 4a is a single 220kV reactive power waveform of the embodiment under PSCAD/EMTDC;

FIG. 4b is a low-voltage side reactive power waveform diagram of a single 110kV main transformer according to the embodiment under PSCAD/EMTDC.

Detailed Description

FIG. 3 shows system wiring and a tidal flow diagram of an embodiment of the present invention: the Guangdong power grid 1 has the typical operation condition of 4 500kV transformer substations with 1000MVA main transformers.

The short-circuit current of the 500kV power supply bus is 40kA, and the requirement of short-circuit current control of the Guangdong power grid is met. The main transformer load factor is about 75%. The 500kV main transformer supplies power to 6 220kV main transformers through 3 identical 220kV lines. The length of each circuit of 220kV line is 50km, the power transmission flow is about 250MW, 2 220kV transformers are supplied with power, and the operation condition of 1 circuit of 220kV line is shown in detail in the figure. The system load flow calculation result is shown in fig. 3, the node voltages are all in a reasonable range, the load flow transmitted by each loop is reasonable, and the power factor meets the relevant operation and management requirements.

The power of the high-voltage side of the 500kV main transformer is 735.5 MW. In the comprehensive load of a 220kV main transformer at 10kV side, the large industrial load is 4.763MW, and the air-conditioning load is 4.818 MW. The 110kV main transformer 10kV side comprehensive load medium and large industrial load is 18.62MW, and the air conditioner load is 16.91 MW. The rest is constant impedance load.

In all the loads of 10kV voltage class of 1 500kV main transformer, the load of an induction motor is (4.763+4.818+18.62+16.91) × 6 is 270.666MW, and the load proportion K of the induction motor_P270.666/735.5 × 100% ═ 36.8%, where the large industrial load is 140.298MW, accounting for 52%; the air conditioning load is 130.368MW, accounting for 48%. The motor parameters are shown in table 1.

TABLE 1 Large Industrial Motor and air conditioner Integrated Motor parameters

Type (B)	R_s	X_s	X_M	R_R	X_R	Tj	Load factor k_L
								Large-scale industrial motor	0.013	0.067	3.8	0.009	0.17	3	0.8
Air conditioner integrated motor	0.064	0.091	2.23	0.059	0.071	0.68	0.8

And then, estimating the maximum dynamic reactive power requirement of the total station comprehensive load when the three-phase metallic short circuit is generated on the high-voltage side of the 500kV main transformer by adopting the algorithm.

The basic data and the specific numerical values required in the method are as follows:

1) the number n of main transformer stations of the transformer substation is 4 (stations) and the short-circuit current level I of the high-voltage bus_d＝40kA(kA)；

2) Active power ratio K of induction motor in transformer substation main transformer high-voltage side with active power P being 735.5(MW) and comprehensive load in normal operation_P＝36.8％；

3) Inertia time constant T of induction motor_jLoad factor K_LAnd equivalent electric road single parameter: stator resistance R_sStator reactance X_sExcitation reactance X_MRotor reactance X_RRotor resistance R_RSee table 1.

4) Failure duration Δ t is 0.1s

The method comprises the following steps:

as the running conditions of 4 main transformers are completely consistent, the following calculation data of 1 main transformer are given, and the result multiplied by 4 is the maximum dynamic reactive power requirement of the total station comprehensive load.

1) Running data calculation under steady state conditions

When the system normally operates, the main transformer input power P is 735.5MW, the comprehensive load power factor is 0.95, and the total power of the comprehensive load is as follows: p + jQ ═ (735.5+ j241.75) MVA. By formula (6), the equivalent impedance of the integrated load of 1 500kV main transformer is as follows:

1.2 Induction Motor equivalent impedance per Unit value calculation

For large industrial motors:

obtained by the formula (2), X_RM＝X_R+X_M＝3.97

X_SM＝X_S+X_M＝3.867

X_P＝X_SX_R+X_SX_M+X_MX_R＝0.912

Is obtained by the formula (4),

A = R_{S}^{2} + X_{SM}^{2} = 14.954

B = 2 R_{S} X_{M}^{2} = 0.37544

C = X_{P}^{2} + R_{S}^{2} X_{RM}^{2} = 0.8344

take V ═ 1.0p.u., P ═ K_L0.8, obtained from formula (7),

A^{'} = \frac{AP}{V^{2}} - R_{s} = 11.9502

B^{'} = \frac{BP}{V^{2}} - X_{M}^{2} = - 14.14

C^{'} = \frac{CP}{V^{2}} - R_{s} X_{RM}^{2} = 0.4626

is obtained by the formula (8),

<math> <mrow> <msub> <mi>R</mi> <mrow> <mi>g</mi> <mn>0</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>-</mo> <msup> <mi>B</mi> <mo>'</mo> </msup> <mo>&PlusMinus;</mo> <msqrt> <msup> <mi>B</mi> <mrow> <mo>'</mo> <mn>2</mn> </mrow> </msup> <mo>-</mo> <mn>4</mn> <msup> <mi>A</mi> <mo>'</mo> </msup> <msup> <mi>C</mi> <mo>'</mo> </msup> </msqrt> </mrow> <mrow> <mn>2</mn> <msup> <mi>A</mi> <mo>'</mo> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1.1496</mn> </mrow> </math>

the initial slip of the large industrial motor obtained by the formula (9) is as follows:

substituted into formula (1):

the initial slip of the air conditioner comprehensive motor can be obtained in the same way: r_k0＝1.096，s_k0＝0.0538；

1.3 constant impedance load equivalent impedance Z_Z

Total load power of induction motor is P₁＝K_PX P36.8% x 735.5 270.7 MW. The large industrial load accounts for 52%, and is 270.7 × 52% — 140.8 MW. The air conditioning load is 48%, and is 270.7 × 48% ═ 129.9 MW. Calculating in step 1.2 to obtain the equivalent impedance angle of the industrial load induction motor of 27.42 degrees and the equivalent impedance angle of the air-conditioning load induction motor of 31.78 degrees

Then: power consumption by industrial load: s_g＝(140.8+j73.05)MVA

Air conditioning load power consumption: s_k＝(129.9+j80.5)MVA

Constant impedance load dissipation power:

S_Z＝(735.5+j241.75)-(140.8+j73.05)-(129.9+j80.5)

＝(464.8+j88.2)MVA

the constant impedance load equivalent impedance is obtained by equation (6):

2) calculating related parameters of the equivalent circuit of the power supply network of the 500kV transformer substation:

equivalent reactance behind 500kV generating line:

and (3) returning the power supply behind the 500kV bus to the post per unit transformation ratio E:

3) calculation of relevant parameters at fault removal time

Large industrial motor speed variation:

induction motor slip at the time of fault removal: s ═ s₀- Δ ω ═ 0.00783+0.0267 ═ 0.03453, equivalent resistance of the induction motor rotor at the moment of fault removal:

the change of the rotating speed of the air conditioner comprehensive motor:induction motor slip at the end of the fault: s is 0.0538+0.1176 is 0.1714, and the equivalent resistance of the induction motor rotor at the end of the fault:

R_{k}^{'} = \frac{0.059}{0.1714} = 0.3442

power reference value of large industrial load induction motor:

air conditioner integrated load induction motor power reference value:

equivalent impedance of large industrial motor:

equivalent impedance of the air conditioner integrated motor:

comprehensive load impedance:

Z_Σ＝Z'_g//Z'_k//Z_Z＝549.68∠44.35°//736.7∠28.583°//582.6∠10.74°

＝211∠28.23°Ω

4) 10kV side bus voltage at fault removal time:

5) estimating the maximum dynamic reactive power demand of the 500kV main transformer comprehensive load during the main network fault:

the dynamic reactive power requirement of the large-scale industrial motor is obtained by the following formula (5):

the dynamic reactive power requirement of the air conditioner comprehensive motor is as follows: q_k＝0.795²×1.105×162＝113Mvar

Total dynamic reactive power demand maximum: q_Σ＝Q_g+Q_k＝334.5Mvar

Fig. 4 is a reactive power waveform diagram of a single 220kV and 110kV main transformer low-voltage side of a specific embodiment system under PSCAD/EMTDC, which is a reactive power waveform diagram of a 220kV and 110kV main transformer low-voltage side obtained by simulation, and 1 500kV main transformer with 6 220kV and 18 110kV main transformers operates, and the total dynamic reactive power demand maximum value: q_∑＝6×4+18×16.5＝321MVar。

The estimation result is well matched with the simulation result.

Claims

1. A method for quickly estimating the maximum reactive power demand of a 500kV radial power supply network load in disturbance recovery comprises the following basic data:

the number n of main transformers of a 1,500 kV transformer substation and the short-circuit current level Id of a high-voltage bus in a unit kA;

2, the active power of the induction motor in the high-voltage side P, unit MW and the comprehensive load of the transformer substation during normal operation_P；

3, inertia time constant T of induction motor_jLoad factor K_LAnd equivalent electric road single parameter: statorResistance R_sStator reactance X_sExcitation reactance X_MRotor reactance X_RRotor resistance R_R；

4, fault duration Δ t;

the method requires the use of the following basic formula:

1) basic formula of induction motor

The induction motor has the equivalent of: stator resistors R connected in series in sequence_sStator reactance X_sRotor reactance X_RAnd a variable resistance R ═ R_RS, in rotor reactance X_RAnd a variable resistance R ═ R_RThe two ends of/s are provided with excitation reactance X_M；

The equivalent impedance Z of the induction motor is calculated as formula (1):

Z = R_{S} + {jX}_{S} + \frac{({jX}_{M}) (R + {jX}_{R})}{R + j (X_{M} + X_{R})} - - - (1);

the defining variables are as in formula (2):

\{\begin{matrix} X_{RM} = X_{R} + X_{M} \\ X_{SM} = X_{S} + X_{M} \\ X_{P} = X_{S} X_{R} + X_{S} S_{M} + X_{M} X_{R} \end{matrix} - - - (2);

\{\begin{matrix} Y = \frac{1}{Z} = \frac{R + {jX}_{RM}}{(R_{S} R - X_{P}) + j (R_{S} X_{RM} + {RX}_{SM})} \\ = \frac{(R_{S} R^{2} + X_{M}^{2} R + R_{S} X_{RM}^{2}) - j (X_{SM} R^{2} + X_{P} X_{RM})}{{AR}^{2} + BR + C} \end{matrix} - - - (3);

wherein,

\{\begin{matrix} A = R_{S}^{2} + X_{SM}^{2} \\ B = 2 R_{S} X_{M}^{2} \\ C = X_{P}^{2} + R_{S}^{2} X_{RM}^{2} \end{matrix} - - - (4);

\{\begin{matrix} P = REAL (V^{2} \overset{*}{Y}) = \frac{V^{2} (R_{S} R^{2} + X_{M}^{2} R + R_{S} X_{RM}^{2})}{{AR}^{2} + BR + C} \\ Q = IMAG (V^{2} \overset{*}{Y}) = \frac{V^{2} (X_{SM} R^{2} + X_{P} X_{RM})}{{AR}^{2} + BR + C} \end{matrix} - - - (5);

2) equivalent impedance calculation formula of electrical element

\{\begin{matrix} r = \frac{P}{P^{2} + Q^{2}} V^{2} \\ x = \frac{Q}{P^{2} {+ Q}^{2}} V^{2} \end{matrix} - - - (6);

the method is characterized by comprising the following steps:

s1, calculating running data under steady state conditions

S1.1 main transformer comprehensive load equivalent impedance calculation

According to the active power P on the high-voltage side of the main transformer of the transformer substation in normal operation in basic data, the unit MW is considered, P is the active power consumed by the comprehensive load on the 10kV side of the main transformer, the power factor is 0.95, and the reactive power Q (cos) consumed by the comprehensive load on the 10kV side of the main transformer with the parallel capacitance is obtained^-1(0.95)); and (3) obtaining the comprehensive load equivalent impedance Z on the 10kV side of the main transformer according to the formula (6) by taking the comprehensive load operating voltage V as 1p.u_Σ0；

S1.2 Induction Motor equivalent impedance per unit value calculation

Taking the terminal voltage V of an induction motor as 1p.u., and taking the active power P as K_Lp.u., according to formulae (2) and (4) and having:

<math> <mrow> <msub> <mi>R</mi> <mn>0</mn> </msub> <mo>=</mo> <mfrac> <mrow> <mo>-</mo> <msup> <mi>B</mi> <mo>′</mo> </msup> <mo>&PlusMinus;</mo> <msqrt> <msup> <mi>B</mi> <mrow> <mo>′</mo> <mn>2</mn> </mrow> </msup> <mo>-</mo> <mn>4</mn> <msup> <mi>A</mi> <mo>′</mo> </msup> <msup> <mi>C</mi> <mo>′</mo> </msup> </msqrt> </mrow> <mrow> <mn>2</mn> <msup> <mi>A</mi> <mo>′</mo> </msup> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>

initial slip of induction motor:

s_{0} = \frac{R_{R}}{R_{0}} - - - (9);

S1.3 constant impedance load equivalent impedance Z_Z

Subtracting the power consumed by the induction motor from the total load power of the main transformer to obtain the power consumed by the constant impedance load, and then obtaining the equivalent impedance Z of the constant impedance load by using the step (6)_Z；

S2, calculating related parameters of the equivalent circuit of the power supply network of the 500kV transformer substation;

s3, calculating related parameters at the fault removal time;

s5, estimating the maximum dynamic reactive power demand of the 500kV main transformer comprehensive load when the main network fails

2. The method for fast estimating the maximum reactive power demand of the load of the 500kV radial power supply network in disturbance recovery as claimed in claim 1, wherein: the step S2 specifically includes: let the general form of a 500kV substation power supply network be: the 500kV transformer substation supplies power to the 220kV transformer substation, and a comprehensive load is connected to the 10kV side of the 220kV transformer substation; the 220kV transformer substation supplies power to the 110kV transformer substation, and a comprehensive load is connected to the 10kV side of the 110kV transformer substation;

the general form is equivalent to: the main network power supply is connected with the power supply through a transformation ratio of 1: e equivalent reactance Xs of an ideal transformer series system, and equivalent impedance X of a power transmission loop to a comprehensive load Z_Σ0The form of the power supply;

the system equivalent reactance Xs is estimated according to equation (10):

3. root of herbaceous plantThe method for fast estimating the maximum reactive power demand of the load of the 500kV radial power supply network in disturbance recovery as claimed in claim 1, wherein: the step S3 specifically includes: induction motor terminal voltage during fault is considered to be zero, induction motor electromagnetic torque T_eInduction motor load torque T of 0_load＝K_LAnd delta t is the fault duration, and the induction motor rotating speed change delta omega at the fault removal moment is obtained according to the step (12):

s＝s₀-Δω (13)；

induction motor rotor resistance at the time of failure completion:

R = \frac{R_{R}}{s} - - - (14);

reference value of power of induction motor:

Z_Σ＝Π(Z) (16)；

II represents parallel connection;

s4, estimating the voltage of the bus at the 10kV side of the 220kV and 110kV transformer substations at the fault removal time