电力系统分析何仰赞版 第七、八章

1、7-3系统电线图如图7-3,各元件参数如下。发电机 G : Sn = 30 MVA ,七=x=0.2;变压器T - 1: Sn = 30MVA ,Vs = 10.5%,中心点接地电阻z = J10;变压器 T - 2 : Sn = 30MVA ,Vs = 10.5%;线路 L : l = 60 km , x = 0.4Q / km , x = 3x ;负荷：S = 25 MVA。计算各元件电抗的标幺值，并作出各序网络。解：求各元件的参数标幺值选 S = 30 MVA , V = V发电机 G : x = x = x = 0.2;静止元件的正、负序电抗相等：变压器 T - 1: x 1（1） =

2、 x 1（2） = 0.105 ,变压器 T - 2: x 2=x 2=0.105 ;S 八，30线路 L : x . = x = xL b- = 60 x 0.4 x= 0.0544B对于负荷:S30 x = 1.2= 1.2 x 一 = 1.44LDx = 0.35 BLD （2）SLD=0.35 x 30 = 0.4225正序网络负序网络零序网络7-5电力系统接线如图75 f发生接地短路作出系统的零序等值网络。题图中117为原件编号解：8-1简单系统如题图8-1所示。已知元件参数如下。发电机 G : S = 60MVA ,x = 0.16,x = 0.19;变压器 T : S = 60M

3、VA ,V = 10.5%,在f点分别发生单相接地、两相短路、两相短路接地和三相短路时，试计算短路点短路电流的有名值， 并进行比较分析。题图8-1选 S = 60 MVA , V = V , E = 1.05,则/SB60V3 x 115kA = 0.301226 kAx=x=0.16G (1)dx=0.19G (2)x=x= x = 0.105T (1)T (2)T (0)所以x=x + x = 0.16 + 0.105 = 0.265解：ff (1) G (1) T (1)x f(2)= XG(2)+ XT(2)= 0.19 + 0.105 = 0.295 x.= x = 0.1053 E

4、3 x 1.05I =1 =x 0.301226 kA = 1.42686 kAf(1) _ f(2) f(0)_v3Ey3 x 1.05I =1 =x 0.301226 kA = 0.9783 kAff (1) ff (2)x (1,1)AX ff (2) X ff (0)0.295 x 0.105 =0.07744x# (2) + x ff (0)0.295 + 0.105m(1,1) = ： 3 x ,1 一 -X f (2) x# (0)-0.295 x 0.105v3 x -1 (0.295 + 0.105 )2=1.5554I (1,1) = f1.5554 x 1.05 x 0.

5、301226 kA = 1.437 kAEm (1,1)1x (1,1) b 0.265 + 0.07744AE 1.05I (3)=1 =x 0.301226 kA = 1.194 kAff (1)由本题可知： I (】) I I (3) I (2)。f f f f8-2上题系统中，若变压器中性点经30欧姆的电抗接地，做题所列各类短路的计算，并对两题计算的结果做分析比较。BBavBx= x = 0.16xc (2) = 0.19七=七=七(0) = 0.105X = 30 七=30 x -6 = 0.13611V2115 2B所以 x .= x + x = 0.16 + 0.105= 0.2

6、65x.= x + x = 0.19 + 0.105= 0.295x f (0) = xT(0) + 3X = 0.105 + 3 x 0.13611 =0.5133中性点接地电阻只对接地故障有影响，故重算接地短路电流解：选 S = 60 MVA , V = V , E = 1.05,则/S*3VB60 kA = 0.301226 kA 3 x 115， + X.ff (2)IB 0.2653 x 1.05+ 0.295 + 0.5133x 0.301226 kA = 0.884 kA=0.9783 kAx (1,1) =*#(2) *#(0)A0.295 x 0.51330.295 + 0.

7、5133m (1,1)=心 x 1 -Gff (2)+尤并(0)=0.1873r I=3x ，1 一0.295 x 0.5133=1.5181(0.295 + 0.5133 )2Em (1,1)f = x + x(1,1) Ibff (1) A1.5181 x 1.05 x 0.301226 kA = 1.062 kA0.265 + 0.1873I (3) = 1.194 kA f中性点接地电抗使零序电抗增大很多，使接地电流大大减小。8-3简单系统如题图8-3所示。已知元们参数如下。发电机 G : S = 50 MVA ,七=x = 0.2, E = 1.05;变压器T : SN = 50 M

8、VA , V = 10.5%, Y, d 11接法，中心点接地电抗22。;在f点分别发生bc两相短路，计算短路点各相电流，并画出bc两相的电流、电压向量图。题图8-3解：选 S = 50 MVA , V = V , Eav 0=1.05 / 90 0,则/B50 :kA = 0.25102 kA3VBu3 x 115x= x = 0.2x= 0.2XT (1) = XT (2) = XT (0) = 0,105X = 22 ； = 22 x -5 = 0.08318 V2115 2B所以 xff =XG + XT=0.2 + 0.105 = 0.305x f(2)= xG(2)+ xT(2)=

9、 0.2 + 0.105 = 0.305x.,(0) = 乂了(0) + 3X = 0.105 + 3 x 0.08318 = 0.35453两相接地短路电流IfaEE0dj ( xf (1) + xf (2)儿05j (0.305 + 0.305 )1.721 , a (2) =Tfa (1) =- 1.721Ifa (0)=0所以Ifa, , =1 fa + 1 fa + 1 fa (0)T721T721I = (I + I + I ) I fbfb (1) fb (2) fb (0) B =(I- 1Z(-120 0) + I- 1Z120 0 + I - 1Z00) x 0.25102 kAfa (1、fa (2)fa (0)=-八；3 I x 0.25102 kA=0.748 /(-90 0)kA. I = -1fc fb=0.748 /90 0 kAV =(V + V + V)vfafa (1) fa (