CN104899677B - The radial supply network load maximum reactive requirements of 500kV estimate method soon during disturbance restores - Google Patents
The radial supply network load maximum reactive requirements of 500kV estimate method soon during disturbance restores Download PDFInfo
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Abstract
Description
技术领域technical field
本发明涉及一种扰动恢复中500kV辐射状供电网络负荷最大无功需求快估方法,具体指一种主网故障后500kV变电站辐射状供电网络综合负荷最大无功需求(近似认为发生在故障切除时刻)的快速估算方法。The present invention relates to a method for quickly estimating the maximum reactive power demand of a 500kV radial power supply network load during disturbance recovery, specifically referring to a method for estimating the maximum reactive power demand of a comprehensive load of a 500kV substation radial power supply network after a main network failure (approximately considered to occur at the time of fault removal) ) for a quick estimate.
背景技术Background technique
电力系统分析计算中,常将电力网覆盖的广大地区内难以胜数的电力用户合并为数量不多的综合负荷,分接在变电站不同电压等级的母线上。电力系统电压稳定性研究中,综合负荷常被表示为不同比例的恒定阻抗和感应电动机的组合。In the analysis and calculation of the power system, the innumerable power users in the vast area covered by the power grid are often combined into a small number of comprehensive loads, which are tapped on the buses of different voltage levels in the substation. In the study of power system voltage stability, the integrated load is often represented as a combination of constant impedance and induction motors with different proportions.
主变高压侧发生短路故障时,电力系统继电保护装置会迅速切除故障。故障后恢复过程中,综合负荷会产生显著的动态无功需求,其大小取决于故障类型、故障持续时间、负荷特性和系统电压恢复;而系统电压恢复又受交流系统强度和负荷动态无功功率需求的影响,二者相互耦合。对此综合负荷的动态无功需求,需借助电力系统计算分析软件才能实现较为精确的计算,比较费时费力。When a short-circuit fault occurs on the high-voltage side of the main transformer, the power system relay protection device will quickly remove the fault. In the recovery process after a fault, the integrated load will generate a significant dynamic reactive power demand, and its size depends on the fault type, fault duration, load characteristics and system voltage recovery; and the system voltage recovery is affected by the AC system strength and load dynamic reactive power The impact of demand, the two are coupled with each other. For the dynamic reactive power demand of the comprehensive load, it is necessary to use power system calculation and analysis software to achieve a more accurate calculation, which is time-consuming and laborious.
发明内容Contents of the invention
本发明所要解决的技术问题,就是提供一种扰动恢复中500kV辐射状供电网络负荷最大无功需求快估方法,所针对的故障类型为电力系统最严重的三相金属性接地短路,计算所需基础数据少、速度快、省时省力。The technical problem to be solved by the present invention is to provide a method for quickly estimating the maximum reactive power demand of a 500kV radial power supply network in disturbance recovery. Less basic data, fast speed, save time and effort.
解决上述技术问题,本发明采用的技术方案是:To solve the problems of the technologies described above, the technical solution adopted in the present invention is:
一种扰动恢复中500kV辐射状供电网络负荷最大无功需求快估方法,方法中所需基础数据有:A fast estimation method for the maximum reactive power demand of a 500kV radial power supply network in disturbance recovery. The basic data required in the method are:
1)500kV变电站主变台数n和高压母线短路电流水平Id,单位kA;1) The number n of main transformers in the 500kV substation and the short-circuit current level Id of the high-voltage bus, in kA;
2)正常运行时变电站主变高压侧有功P,单位MW,以及综合负荷中感应电动机的有功占比KP;2) During normal operation, the active power P of the high-voltage side of the main transformer of the substation, unit MW, and the active power ratio K P of the induction motor in the comprehensive load;
3)感应电动机惯性时间常数Tj,负载率KL,等值电路标幺参数:定子电阻Rs,定子电抗Xs,励磁电抗XM,转子电抗XR,转子电阻RR;3) Inertia time constant T j of induction motor, load rate K L , equivalent circuit unit parameters: stator resistance R s , stator reactance X s , excitation reactance X M , rotor reactance X R , rotor resistance R R ;
4)故障持续时间Δt;4) Fault duration Δt;
所述方法需要使用以下基本公式:The method described requires the use of the following basic formula:
1)感应电动机基本公式1) Basic formula of induction motor
参见图1,感应电动机等值为:依次串联的定子电阻Rs、定子电抗Xs、转子电抗XR和可变电阻R=RR/s,在转子电抗XR和可变电阻R=RR/s两端设有励磁电抗XM;Referring to Figure 1, the equivalent value of the induction motor is: the stator resistance R s, the stator reactance X s, the rotor reactance X R and the variable resistance R=R R /s in series, and the rotor reactance X R and the variable resistance R=R There is an exciting reactance X M at both ends of R /s;
感应电动机等值阻抗Z计算如式(1):The equivalent impedance Z of the induction motor is calculated as formula (1):
定义变量如式(2):Define variables as formula (2):
则感应电动机等值导纳Y计算如式(3):Then the equivalent admittance Y of the induction motor is calculated as formula (3):
其中, in,
感应电动机运行电压为V时,消耗的功率P+jQ计算如式(5):When the operating voltage of the induction motor is V, the power consumption P+jQ is calculated as formula (5):
2)电气元件等值阻抗计算公式2) Calculation formula of equivalent impedance of electrical components
若电气元件运行电压为V,消耗的功率为P+jQ,则该电气元件的等值阻抗Z=r+jx计算如式(6):If the operating voltage of the electrical component is V and the power consumed is P+jQ, then the equivalent impedance Z=r+jx of the electrical component is calculated as formula (6):
所述方法包括以下步骤:The method comprises the steps of:
1)稳态条件下运行数据计算1) Calculation of operating data under steady state conditions
1.1主变综合负荷等值阻抗计算1.1 Calculation of the equivalent impedance of the main transformer comprehensive load
根据基础数据中正常运行时变电站主变高压侧有功P,单位MW,并认为P即为主变10kV侧综合负荷消耗的有功功率,功率因数为0.95,得已计及并联电容的主变10kV侧综合负荷消耗的无功Q=P·tg(cos-1(0.95))。取综合负荷运行电压V=1p.u.,根据式(6)求得主变10kV侧综合负荷等值阻抗ZΣ0;According to the basic data in the normal operation of the substation main transformer high-voltage side active power P, the unit is MW, and it is considered that P is the active power consumed by the comprehensive load on the 10kV side of the main transformer, and the power factor is 0.95, which has taken into account the parallel capacitance of the main transformer 10kV side Reactive power Q=P·tg(cos -1 (0.95)) consumed by integrated load. Take the comprehensive load operating voltage V=1p.u., and obtain the comprehensive load equivalent impedance Z Σ0 of the 10kV side of the main transformer according to formula (6);
1.2感应电动机等值阻抗标幺值计算1.2 Calculation of equivalent impedance per unit value of induction motor
取感应电动机端电压V=1p.u.,有功功率P=KL p.u.,根据式(2)和式(4),并令:Take induction motor terminal voltage V=1p.u., active power P=K L pu, according to formula (2) and formula (4), and make:
求得正常运行时感应电动机转子等值电阻:Obtain the equivalent resistance of the induction motor rotor during normal operation:
感应电动机初始转差: Induction motor initial slip:
由式(1)计算感应电动机等值阻抗标幺值ZM*;Calculate the per unit value Z M* of the equivalent impedance of the induction motor by formula (1);
1.3恒阻抗负荷等值阻抗ZZ 1.3 Constant impedance load equivalent impedance Z Z
根据基础数据中给出的正常运行时变电站主变高压侧有功P,单位MW,及综合负荷中感应电动机的有功占比KP,知KP·P为电动机消耗的有功功率,根据ZM*对应的功率因数角,求得电动机消耗的无功 According to the active power P of the high-voltage side of the main transformer substation given in the basic data during normal operation, the unit is MW, and the active power ratio of the induction motor in the comprehensive load K P , it is known that K P P is the active power consumed by the motor, according to Z M* Corresponding power factor angle, get the reactive power consumed by the motor
从主变总负荷功率中减去感应电动机消耗的功率,得恒阻抗负荷消耗的功率,再利用(6)求得恒阻抗负荷等值阻抗ZZ;Subtract 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 use (6) to obtain the equivalent impedance Z Z of the constant impedance load;
2)500kV变电站供电网络等值电路相关参数计算2) Calculation of relevant parameters of equivalent circuit of 500kV substation power supply network
500kV变电站供电网络的通常形式见图2a:500kV变电站向220kV变电站供电,220kV变电站10kV侧接有综合负荷;220kV变电站再向110kV变电站供电,110kV变电站10kV侧接有综合负荷;将图2a等值为图2b:主网供电电源经过变比1:E的理想变压器串联系统等值电抗Xs,经输电回路等值阻抗X向综合负荷ZΣ0供电的形式;The general form of the 500kV substation power supply network is shown in Figure 2a: the 500kV substation supplies power to the 220kV substation, and the 10kV side of the 220kV substation is connected to the integrated load; the 220kV substation supplies power to the 110kV substation, and the 10kV side of the 110kV substation is connected to the integrated load; Figure 2b: The power supply of the main network passes through the equivalent reactance Xs of the ideal transformer series system with a transformation ratio of 1:E, and supplies power to the comprehensive load Z Σ0 through the equivalent impedance X of the transmission circuit;
按式(10)估算系统等值电抗Xs:Estimate system equivalent reactance Xs according to formula (10):
综合负荷与500kV主变高压侧的电气距离按X=140Ω估算;The electrical distance between the integrated load and the high voltage side of the 500kV main transformer is estimated by X=140Ω;
按照正常运行时综合负荷端电压为1p.u.估算供电回路标幺变比E如式(11):According to the comprehensive load terminal voltage during normal operation is 1p.u. Estimate the per-unit transformation ratio E of the power supply circuit as formula (11):
3)故障切除时刻相关参数计算3) Calculation of relevant parameters at the time of fault removal
故障期间感应电动机端电压考虑为零,感应电动机电磁转矩Te=0,感应电动机负载转矩Tload=KL,Δt为故障持续时间,据(12)求得故障切除时刻感应电动机转速变化Δω:During the fault period, the terminal voltage of the induction motor is considered to be zero, the electromagnetic torque of the induction motor T e = 0, the load torque of the induction motor T load = K L , Δt is the duration of the fault, and the speed change of the induction motor at the moment of fault removal is obtained according to (12) Δω:
根据式(13)求得故障结束时刻感应电动机转差:Calculate the slip of the induction motor at the end of the fault according to formula (13):
s=s0-Δω (13);s=s 0 -Δω (13);
故障结束时刻感应电动机转子电阻:The rotor resistance of the induction motor at the end of the fault:
感应电动机功率基准值:Induction motor power reference:
由式(1)计算故障切除时刻综合负荷中各类感应电动机负荷的等值阻抗;综合负荷等值阻抗ZΣ为感应电动机和恒阻抗负荷之并联值,计算如式(16):Calculate the equivalent impedance of various induction motor loads in the comprehensive load at the time of fault removal by formula (1); the comprehensive load equivalent impedance Z Σ is the parallel connection value of the induction motor and the constant impedance load, calculated as formula (16):
Z∑=∏(Z) (16);Z ∑ = ∏(Z) (16);
∏表示并联;Π means parallel connection;
4)故障切除时刻,220kV和110kV变电站10kV侧母线电压估算4) At the time of fault removal, the bus voltage estimation on the 10kV side of 220kV and 110kV substations
5)主网故障时500kV主变综合负荷最大动态无功需求估算5) Estimation of the maximum dynamic reactive power demand of the 500kV main transformer comprehensive load when the main grid fails
将U(10)代入(5)式中无功计算式,得500kV主变综合负荷中各种感应电动机动态无功最大值,求和,即为主网故障时500kV主变综合负荷最大动态无功需求。Substituting U (10) into the reactive power calculation formula in (5), the maximum value of dynamic reactive power of various induction motors in the comprehensive load of the 500kV main transformer can be obtained. work needs.
备注:1)综合负荷中大约20%由220kV变电站低压侧供电,这部分负荷距离500kV主变高压侧电气距离较近,而算法中全部负荷按照110kV变电站低压侧供电考虑,这会造成估算的综合负荷最大动态无功需求偏低;2)故障切除时刻,负荷母线电压偏低,感应电动机以外的负荷和并联电容等值阻抗取用无功也会相应降低,算法中忽略了其影响,会造成估算的综合负荷最大动态无功需求偏高;两方面的影响基本抵消。Remarks: 1) About 20% of the comprehensive load is powered by the low-voltage side of the 220kV substation. This part of the load is relatively close to the high-voltage side of the 500kV main transformer. In the algorithm, all loads are considered as the power supply of the low-voltage side of the 110kV substation, which will result in a comprehensive estimate. The maximum dynamic reactive power demand of the load is low; 2) When the fault is removed, the load bus voltage is low, and the reactive power taken by the load other than the induction motor and the equivalent impedance of the parallel capacitor will also decrease accordingly. The algorithm ignores its influence, which will cause The estimated maximum dynamic reactive power demand of integrated load is on the high side; the effects of the two aspects are basically offset.
有益效果:一般采取仿真手段求得扰动恢复中500kV辐射状供电网络负荷最大无功需求,费时费力,目前未见相关解析计算方法报道。本发明计算所需基础数据少、速度快、省时省力。Beneficial effects: Generally, the simulation method is used to obtain the maximum reactive power demand of the 500kV radial power supply network load in the disturbance recovery, which is time-consuming and labor-intensive, and there is no report on the relevant analytical calculation method at present. The invention requires less basic data for calculation, is fast, and saves time and effort.
附图说明Description of drawings
图1为感应电动机等值电路图;Figure 1 is an equivalent circuit diagram of an induction motor;
图2a为通常500kV变电站供电网络图;Figure 2a is a typical 500kV substation power supply network diagram;
图2b为通常500kV变电站供电网络图之等值电路;Figure 2b is the equivalent circuit of a typical 500kV substation power supply network diagram;
图3为具体实施算例系统接线及潮流图;Figure 3 is the system wiring and power flow diagram of the specific implementation example;
图4a为PSCAD/EMTDC下具体实施例的单台220kV无功波形图;Fig. 4 a is the single 220kV reactive waveform diagram of the specific embodiment under PSCAD/EMTDC;
图4b为PSCAD/EMTDC下具体实施例的单台110kV主变低压侧无功波形图。Fig. 4b is a reactive power waveform diagram of a single 110kV main transformer low-voltage side of a specific embodiment under PSCAD/EMTDC.
具体实施方式Detailed ways
图3给出的是本发明具体实施例的系统接线及潮流图:广东电网1座含4台容量为1000MVA主变的500kV变电站的典型运行情况。Fig. 3 shows the system wiring and power flow diagram of a specific embodiment of the present invention: a typical operation situation of a 500kV substation with four 1000MVA main transformers in Guangdong Power Grid.
500kV供电母线短路电流为40kA,满足广东电网短路电流控制要求。主变负载率约75%。500kV主变经3条相同的220kV线路给6台220kV主变供电。每回220kV线路长50km,送电潮流约250MW,给2台220kV变压器供电,图中详细给出其中1回220kV线路的运行情况。系统潮流计算结果如图3所示,节点电压均在合理范围内,各回线路传输潮流合理,功率因数符合相关运行与管理要求。The short-circuit current of the 500kV power supply busbar is 40kA, which meets the short-circuit current control requirements of Guangdong power grid. The main transformer load rate is about 75%. The 500kV main transformer supplies power to six 220kV main transformers via three identical 220kV lines. Each 220kV line is 50km long, and the transmission flow is about 250MW, which supplies power to two 220kV transformers. The figure shows the operation of one of the 220kV lines in detail. The power flow calculation results of the system are shown in Figure 3. The node voltages are all within a reasonable range, the transmission power flow of each circuit is reasonable, and the power factor meets the relevant operation and management requirements.
500kV主变高压侧的功率为735.5MW。220kV主变10kV侧综合负荷中,大型工业负荷4.763MW,空调负荷4.818MW。110kV主变10kV侧综合负荷中大型工业负荷18.62MW,空调负荷16.91MW。其余为恒阻抗负荷。The power of the high voltage side of the 500kV main transformer is 735.5MW. Among the comprehensive loads on the 10kV side of the 220kV main transformer, the large industrial load is 4.763MW, and the air conditioning load is 4.818MW. The comprehensive load on the 10kV side of the 110kV main transformer is 18.62MW for medium and large industrial loads, and the load for air conditioning is 16.91MW. The rest are constant impedance loads.
1台500kV主变10kV电压等级所有负荷中,感应电动机负荷共(4.763+4.818+18.62+16.91)×6=270.666MW,感应电动机负荷比例KP为270.666/735.5×100%=36.8%,其中大型工业负荷140.298MW,占52%;空调负荷130.368MW,占48%。马达参数见表1。Among all the loads of a 500kV main transformer with 10kV voltage level, the induction motor load is (4.763+4.818+18.62+16.91)×6=270.666MW, and the induction motor load ratio K P is 270.666/735.5×100%=36.8%. The 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.
表1 大型工业马达和空调综合马达参数Table 1 Parameters of large industrial motors and comprehensive motors for air conditioners
下面采用前述的算法估算500kV主变高压侧三相金属性短路时全站综合负荷最大动态无功需求。The above-mentioned algorithm is used to estimate the maximum dynamic reactive power demand of the comprehensive load of the whole station when the three-phase metallic short circuit on the high voltage side of the 500kV main transformer is used.
方法中所需基础数据及其具体数值如下:The basic data and their specific values required in the method are as follows:
1)变电站主变台数n=4(台)和高压母线短路电流水平Id=40kA(kA);1) The number of main transformers in the substation n=4 (sets) and the short-circuit current level of the high-voltage busbar I d =40kA (kA);
2)正常运行时变电站主变高压侧有功P=735.5(MW)及综合负荷中感应电动机的有功占比KP=36.8%;2) During normal operation, the active power of the high-voltage side of the substation main transformer P = 735.5 (MW) and the active power ratio of the induction motor in the comprehensive load K P = 36.8%;
3)感应电动机惯性时间常数Tj,负载率KL,等值电路标幺参数:定子电阻Rs,定子电抗Xs,励磁电抗XM,转子电抗XR,转子电阻RR,见表1。3) Induction motor inertia time constant T j , load rate K L , equivalent circuit unit parameters: stator resistance R s , stator reactance X s , excitation reactance X M , rotor reactance X R , rotor resistance R R , see Table 1 .
4)故障持续时间Δt=0.1s4) Fault duration Δt=0.1s
所述方法包括以下步骤:The method comprises the steps of:
由于4台主变运行情况完全一致,以下给出1台主变的计算数据,结果乘以4即为全站综合负荷最大动态无功需求。Since the operation conditions of the four main transformers are exactly the same, the calculation data of one main transformer is given below, and the result multiplied by 4 is the maximum dynamic reactive power demand of the comprehensive load of the whole station.
1)稳态条件下运行数据计算1) Calculation of operating data under steady state conditions
1.1主变综合负荷等值阻抗计算1.1 Calculation of the equivalent impedance of the main transformer comprehensive load
正常运行时,主变输入功率P=735.5MW,综合负荷功率因数取0.95,则综合负荷总功率为:P+jQ=(735.5+j241.75)MVA。由式(6),1台500kV主变综合负荷等值阻抗为:During normal operation, the input power of the main transformer is P=735.5MW, and the comprehensive load power factor is 0.95, then the total power of the comprehensive load is: P+jQ=(735.5+j241.75)MVA. From formula (6), the comprehensive load equivalent impedance of a 500kV main transformer is:
1.2感应电动机等值阻抗标幺值计算1.2 Calculation of equivalent impedance per unit value of induction motor
对于大型工业马达:For large industrial motors:
由式(2)得,XRM=XR+XM=3.97From formula (2), X RM =X R +X M =3.97
XSM=XS+XM=3.867X SM =X S +X M =3.867
XP=XSXR+XSXM+XMXR=0.912X P =X S X R +X S X M +X M X R =0.912
由式(4)得, From formula (4), we get,
取V=1.0p.u.,P=KL=0.8,由式(7)得,Take V=1.0pu, P=K L =0.8, get from formula (7),
由式(8)得, From formula (8), we get,
由式(9)得大型工业马达初始转差为: From formula (9), the initial slip of the large industrial motor is:
代入式(1):Substitution (1):
同理可得空调综合马达初始转差:Rk0=1.096,sk0=0.0538;In the same way, the initial slip of the integrated motor of the air conditioner can be obtained: R k0 = 1.096, s k0 = 0.0538;
1.3恒阻抗负荷等值阻抗ZZ 1.3 Constant impedance load equivalent impedance Z Z
感应电动机负荷总功率为P1=KP×P=36.8%×735.5=270.7MW。其中,大型工业负荷占52%,为270.7×52%=140.8MW。空调负荷占48%,为270.7×48%=129.9MW。计及步骤1.2计算得到工业负荷感应电动机等值阻抗角为27.42°,空调负荷感应电动机等值阻抗角为31.78°The total load power of the induction motor is P 1 =K P ×P=36.8%×735.5=270.7MW. Among them, the large-scale industrial load accounts for 52%, which is 270.7×52%=140.8MW. Air-conditioning load accounts for 48%, which is 270.7×48%=129.9MW. Taking into account the calculation in step 1.2, the equivalent impedance angle of the industrial load induction motor is 27.42°, and the equivalent impedance angle of the air-conditioning load induction motor is 31.78°
则:工业负荷消耗功率:Sg=(140.8+j73.05)MVAThen: industrial load power consumption: S g = (140.8+j73.05) MVA
空调负荷消耗功率:Sk=(129.9+j80.5)MVAAir conditioning load power consumption: S k = (129.9+j80.5)MVA
恒阻抗负荷消耗功率:Constant impedance load power consumption:
SZ=(735.5+j241.75)-(140.8+j73.05)-(129.9+j80.5)S Z =(735.5+j241.75)-(140.8+j73.05)-(129.9+j80.5)
=(464.8+j88.2)MVA=(464.8+j88.2)MVA
由式(6)得,恒阻抗负荷等值阻抗:From formula (6), the equivalent impedance of constant impedance load is:
2)500kV变电站供电网络等值电路相关参数计算:2) Calculation of relevant parameters of equivalent circuit of 500kV substation power supply network:
500kV母线背后等值电抗:Equivalent reactance behind the 500kV busbar:
500kV母线背后供电回路标幺变比E:The power supply circuit behind the 500kV bus bar has a per-unit transformation ratio E:
3)故障切除时刻相关参数计算3) Calculation of relevant parameters at the time of fault removal
大型工业马达转速变化:故障切除时 刻感应电动机滑差:s=s0-Δω=0.00783+0.0267=0.03453,故障切除时刻感应电动机转 子等值电阻: Large industrial motor speed change: Slip of the induction motor at the time of fault removal: s=s 0 -Δω=0.00783+0.0267=0.03453, the equivalent resistance of the induction motor rotor at the time of fault removal:
空调综合马达转速变化:故障结束时刻感应电动 机滑差滑差:s=0.0538+0.1176=0.1714,故障结束时刻感应电动机转子等值电阻: Air conditioner integrated motor speed change: Slip of induction motor at the end of the fault: s=0.0538+0.1176=0.1714, equivalent resistance of the induction motor rotor at the end of the fault:
大型工业负荷感应电动机功率基准值:Large Industrial Duty Induction Motor Power References:
空调综合负荷感应电动机功率基准值:Air conditioner integrated load induction motor power reference value:
大型工业马达等值阻抗:Equivalent impedance of large industrial motors:
空调综合马达等值阻抗:Equivalent impedance of integrated air conditioner motor:
综合负荷阻抗:Integrated Load Impedance:
ZΣ=Z'g//Z'k//ZZ=549.68∠44.35°//736.7∠28.583°//582.6∠10.74°Z Σ =Z' g //Z' k //Z Z =549.68∠44.35°//736.7∠28.583°//582.6∠10.74°
=211∠28.23°Ω=211∠28.23°Ω
4)故障切除时刻10kV侧母线电压:4) The bus voltage on the 10kV side at the time of fault removal:
5)主网故障时500kV主变综合负荷最大动态无功需求估测:5) Estimation of the maximum dynamic reactive power demand of the 500kV main transformer comprehensive load when the main grid fails:
大型工业马达动态无功需求,由式(5)得:The dynamic reactive power demand of large industrial motors can be obtained from formula (5):
空调综合马达动态无功需求:Qk=0.7952×1.105×162=113MvarDynamic reactive power demand of integrated motor of air conditioner: Q k =0.795 2 ×1.105×162=113Mvar
总动态无功需求最大值:QΣ=Qg+Qk=334.5MvarMaximum value of total dynamic reactive power demand: Q Σ =Q g +Q k =334.5Mvar
图4为PSCAD/EMTDC下具体实施算例系统单台220kV和110kV主变低压侧无功波形图,为仿真得到的220kV和110kV主变低压侧无功波形图,1台500kV主变带6台220kV和18台110kV主变运行,总动态无功需求最大值:Q∑=6×4+18×16.5=321MVar。Figure 4 is the reactive power waveform diagram of the low-voltage side of a single 220kV and 110kV main transformer under PSCAD/EMTDC. 220kV and 18 sets of 110kV main transformers are running, the maximum value of total dynamic reactive power demand: Q ∑ =6×4+18×16.5=321MVar.
估算结果与仿真结果吻合良好。The estimated results are in good agreement with the simulation results.
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