鲁东大学电力系统分析A卷答案.doc

一、计算题，共3小题，共40分。1、 某一回110KV架空电力线路，长度为60km，导线型号为LGJ-120，导线计算外径为15.2mm，三相导线水平排列，两相邻导线之间的距离为4m。试计算改电力线路的参数，并作出等值电路。（20分）L=60; S=120;%mm2 r=15.2/2;%mm D12=4;%m D23=4;%m D31=8;%mDM=(D12*D23*D31*(109)(1/3) r1=31.5/S x1=0.1445*log10(DM/r)+0.0157b1=7.58/log10(DM/r)*10(-6)% DM = 5.0397e+003% r1 = 0.2625% x1 = 0.4234% b1 = 2.6864e-006R1=r1*L X1=x1*L B1=b1*L% R1 = 15.7500 % X1 = 25.4052 % B1 = 1.6119e-0042、某枢纽变电所，变压器二次母线要求1011KV，试问当分接头，位置选择为110KV时能否满足调压要求，如不能应如何处理?（10分）Upie_i_max=10;Upie_i_min=11;RT=2.44;%高压侧XT=40;ZT=RT+j*XT;U1N=110;%3个抽头 正负1*2.5%U2N=11;%最大负荷时的数据S2max=28+j*14;P2max=real(S2max);Q2max=imag(S2max);S2maxAbs=abs(S2max);UImax=110;%最小负荷时的数据S2min=10+j*6;P2min=real(S2min);Q2min=imag(S2min);S2minAbs=abs(S2min);UImin=113;%先判断标准分接头是否合理%k1=110/11%不合格 Ui_min_s = 11.0500%k1=(110-2.75)/11%Ui_min_s = 11.3334k1=(110+2.75)/11%Ui_max_s = 10.0591 Ui_min_s = 10.7805 符合1011KVdelta_Smax=(S2maxAbs/U1N)2*ZT%注意要用折算到一次侧的值S1max=S2max+delta_SmaxP1max=real(S1max);Q1max=imag(S1max);delta_Umax=(P1max*RT+Q1max*XT)/UImaxUi_max=UImax-delta_UmaxUi_max_s=Ui_max/k1% k1 = 10% delta_Smax = 0.1976 + 3.2397i% S1max = 28.1976 +17.2397i% delta_Umax = 6.8944% Ui_max = 103.1056% Ui_max_s = 10.3106delta_Smin=(S2minAbs/U1N)2*ZTS1min=S2min+delta_SminP1min=real(S1min);Q1min=imag(S1min);delta_Umin=(P1min*RT+Q1min*XT)/UIminUi_min=UImin-delta_UminUi_min_s=Ui_min/k1% delta_Smin = 0.0274 + 0.4496i% S1min = 10.0274 + 6.4496i% delta_Umin = 2.4996% Ui_min = 110.5004% Ui_min_s = 11.0500UtI_max=Ui_max*U2N/Upie_i_maxUtI_min=Ui_min*U2N/Upie_i_minUtI_av=(UtI_max+UtI_min)/2% UtI_max = 113.4161% UtI_min = 110.5004% UtI_av = 111.9583%所以选+2.5%的抽头 112.75kv3、求图示网络的有功分点、无功分点。线路单位长度参数0.175+j3.6/km。（10分）% 题3-14xiangshi的已知条件如下z1=0.175+j*3.6;%ohm/kml12=17;%kml23=12;l31=16;S2=8+j*5;%MVAS3=10+j*7;%MVA%要求有功和无功分点Zl12=z1*l12Zl23=z1*l23Zl31=z1*l31% Zl12 = 2.9750 +61.2000i% Zl23 = 2.1000 +43.2000i% Zl31 = 2.8000 +57.6000i%根据力矩公式，求功率分点S12=( S3*conj(Zl31) + S2*conj( Zl31 + Zl23 ) )/(conj( Zl31 + Zl23 + Zl12 )S13=( S2*conj(Zl12) + S3*conj( Zl12 + Zl23 ) )/(conj( Zl12 + Zl23 + Zl31 )S23=S12-S2% S12 = 8.5333 + 5.6000i% S13 = 9.4667 + 6.4000i% S23 = 0.5333 + 0.6000i%所以功率分点是节点3，从节点3分开为2个开网二、如图两各独立系统的单位调节功率分别为KA=45MW/Hz，KB=400MW/Hz。并网前A系统的总负荷为PA=1500MVA，频率为49.5Hz；B系统的总负荷为PA=1000MVA，频率为50Hz，计算并网后系统频率及联络线上流过功率。（10分）解： 49.5-f=Pab/45 f-50=Pab/400解得：Pab=-20.225MW、f=49.949Hz法2：delta_Pa=45*0.5=22.5 delta_Pb=0所以 delta_f=-( delta_Pa+ delta_Pb)/(kA+kB)=-0.05Pab=( kA* delta_Pb- kB* delta_Pa ) / (kA+kB )= -20.225F=50-0.05=49.95三、简单系统如图，求图示网络始端电压和输入功率（20分）% 题3-8xiangshi的已知条件如下U2=110;S2=18+j*10;X1=20;R1=5.2;B1half=72*10(-6);%求U1、S1 deltaSY2 = (-1)*j*B1half*(U22)S2pie=S2+deltaSY2delta_S_Z=abs(S2pie)2/(U22)*(R1+j*X1)S1pie=S2pie+delta_S_ZdU=conj(S2pie)/U2*(R1+j*X1)U1=U2+dUdelta_S_Y1=-j*B1half*(abs(U1)2)S1=S1pie+delta_S_Y1U1_abs=abs(U1)U1_angle=angle(U1)*180/pi% deltaSY2 = 0 - 0.8712i% S2pie = 18.0000 + 9.1288i% delta_S_Z = 0.1751 + 0.6733i% S1pie = 18.1751 + 9.8021i% dU = 2.5107 + 2.8412i% U1 = 1.1251e+002 +2.8412e+000i% delta_S_Y1 = 0 - 0.9120i% S1 = 18.1751 + 8.8901i% U1_abs = 112.5466% U1_angle = 1.4466四、在下图所示的简单电路中，C相断开，则,A、B两相电流为，。试以A相电流为参考相量，计算线电流的对称分量。（20分）Iabc=10; -10; 0;a=exp(j*2*pi/3);S=1/3*1 a a2;1 a2 a ;1 1 1;I120=S*IabcI120_abs=abs(I120)I120_degree=180/pi*angle(I120)% I120 =% 5.0000 - 2.8868i% 5.0000 + 2.8868i% 0 % I120_abs =% 5.7735% 5.7735% 0% I120_degree =% -30.0000% 30.0000% 0五、如图所示系统中K点发生A相接地短路（各元件参数示于图中）。求短路点故障相短路电流有名值大小。其中Xp在基准值为SB=100MVA，UB=115kV下标么值为0.1。（10分）% 题8-46xiangshi的已知条件如下SNG=50;X1=0.12;X2=0.12;SNT=50;XT=0.105;UN2=115;SB=100;UB=115;XP_biaoyao=0.1;%求解如下X1SIGMA=X1*2+XT*2X2SIGMA=X2*2+XT*2X0SIGMA=XT*2+XP_biaoyao*3Ia1=1/(X1SIGMA+X2SIGMA+X0SIGMA)Ia=3*Ia1Ia_youmingzhi=Ia*(SB/sqrt(3)/UB)% X1SIGMA = 0.4500% X2SIGMA = 0.4500% X0SIGMA = 0.5100% Ia1 = 0.7092% Ia = 2.1277% Ia_youmingzhi = 1.0682第7页 共8 页 第8页 共8 页