电力系统分析练习题及其答案(何仰赞)上册 下载本文

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3.3 某一额定电压为10kV的两端供电网,如图所示。线路L1、L2和L3导线型号均为LJ-185,线路长度分别为10km,4km和3km,线路L4为2km长的LJ-70导线;各负荷点负

?A荷如图所示。试求V?B?10.4?0?kV时的初始功率分布,?10.5?0?kV、V且找到电压最低点。(线路参数LJ-185:z=0.17+j0.38Ω/km;LJ-70:z=0.45+j0.4Ω/km)

解 线路等值阻抗 ZL1 ZL2 ZL3 ZL4?10?(0.17?j0.38)?1.7?j3.8? ?4?(0.17?j0.38)?0.68?j1.52? ?3?(0.17?j0.38)?0.51?j1.14?

?2?(0.45?j0.4)?0.9?j0.8?

求C点和D点的运算负荷,为

0.32?0.162(0.9?j0.8)?1.04?j0.925kVA ?SCE?210SC?2600?j1600?300?j160?1.04?j0.925

?2901.04?j1760.925kVA

SD?600?j200?1600?j1000?2200?j1200kVA

循环功率

Sc?V??A?10.5?10.4??10?VB??VN??Z?17??0.17?j0.38?

?339.43?0.17?j0.38?kVA?580?j129kVA1SAC??2901.04?7?2200?3?j1760.925?7?j1200?3??Sc17?1582.78?j936.85?580?j129?2162.78?j1065.85kVA1SBD??2901.04?10?2200?14?j1760.925?10?j1200?14??Sc17?3518.26?j2024.07?580?j129?2938.26?j1895.07kVA

验算

SAC?SBD?2162.78?j1065.85?2938.26?j1895.07?5101.04?j2960.92kVASC?SD?2901.04?j1760.925?2200?j1200SDC

?5101.04?j2960.92kVA?SBD?SD?2938.26?j1895.07?2200?j1200

?738.26?j695.07kVAC点为功率分点,可推算出E点为电压最低点。进一步可求得E点电压 ?SAC

2.162?1.072?(1.7?j3.8)MVA?98.78?j220.8kVA 210'SAC?2162.78?j1065.85?98.78?j220.8?2261.56?j1286.65kVA

?VAC2.26?1.7?1.29?3.8??0.8328kV

10.50.301?0.9?0.161?0.8??0.041kV

9.6672VC?VA??VAC?10.5?0.8328?9.6672kV

?VCEVE?VC??VCE?9.6672?0.041?9.6262kV

3.4 图所示110kV闭式电网,A点为某发电厂的高压母线,其运行电压为117kV。网络各组件参数为:

-6

线路Ⅰ、Ⅱ(每公里):r0=0.27Ω,x0=0.423Ω,b0=2.69×10S

-6

线路Ⅲ(每公里):r0=0.45Ω,x0=0.44Ω,b0=2.58×10S 线路Ⅰ长度60km,线路Ⅱ长度50km,线路Ⅲ长度40km 变电所b SN?20MVA,?S0?0.05?j0.6MVA,RT?4.84?,

XT?63.5?

变电所c SN?10MVA,?S0?0.03?j0.35MVA,RT?11.4?,

XT?127?

负荷功率 SLDb?24?j18MVA,SLDc?12?j9MVA

试求电力网络的功率分布及最大电压损耗。

解 (1)计算网络参数及制定等值电路。 线路Ⅰ: Z? B??(0.27?j0.423)?60??16.2?j25.38? ?2.69?10?6?60S?1.61?10?4S

??1.61?10?4?1102Mvar??1.95Mvar

2?QB? 线路Ⅱ: Z? B??(0.27?j0.423)?50??13.5?j21.15? ?2.69?10?6?50S?1.35?10?4S

?422?Q??1.35?10?110Mvar??1.63Mvar B? 线路Ⅱ: Z???? B????(0.45?j0.44)?40??18?j17.6? 2.58?10?6?40S?1.03?10?4S

?1.03?10?4?1102Mvar??1.25Mvar

2?QB????1?4.84?j63.5???2.42?j31.75? 变电所b:ZTb?2 ?S0b?2?0.05?j0.6?MVA?0.1?j1.2MVA

1?11.4?j127???5.7?j63.5? 变电所b:ZTc?2 ?S0c等值电路如图所示

?2?0.03?j0.35?MVA?0.06?j0.7MVA

(2)计算节点b和c的运算负荷。

242?182?2.24?j31.75?MVA?0.18?j2.36MVA ?STb?2110Sb?SLDb??STb??Sob?j?QBI?j?QB????24?j18?0.18?j2.36?0.1?j1.2?j0.975?j0.623MVA?24.28?j19.96MVA122?92?5.7?j63.5?MVA?0.106?j1.18MVA ?STc?2110Sc?SLDc??STc??Soc?j?QB????j?QB???12?j9?0.106?j1.18?0.06?j0.7?j0.623?j0.815MVA?12.17?j9.44MVA(3)计算闭式网络的功率分布。

Sb?Z????Z??????ScZ???S??Z???Z????Z????j19.96??31.5?j38.75???12.17?j9.44??13.5?j21.15??MVA47.7?j64.13?18.64?j15.79MVASc?Z???Z??????SbZ??S???Z???Z????Z??????24.28??12.17?j19.44??34.2?j42.98???24.28?j19.96??16.2?j25.38?47.7?j64.13MVA?17.8?j13.6MVASI?S???18.64?j15.79?17.8?j13.6MVA?36.44?j29.39MVA

Sb?Sc?24.28?j19.96?12.17?j9.44MVA?36.45?j29.4MVA