第六章-短路电流计算习题修改版--案例题

9第八章1短路电流计算习题案例题1.将网络电抗由三角形变成等值星形已知512x，1023x，1531x，求1x，2x，3x123xxxxxx12367.11x，5.22x，53x51x，67.12x，5.23x5.21x，67.12x，53x51x，5.22x，67.13x答案：C计算过程：5.230751510515531231231121xxxxxx67.13010531231223122xxxxxx530151031231231233xxxxxx2.将网络电抗由星形变成等值三角形已知51x，102x，153x，求12x，23x，31x231123xxxxxx33.1812x，5.2723x，5531x5.2712x，33.1823x，5531x5.2712x，5523x，33.1831x33.1812x，5523x，5.2731x答案：D计算过程：1033.18151051053212112xxxxxx555151015101323223xxxxxx5.27105155152131331xxxxxx3.将叉形（两叉）网络变成等值非叉形已知51x，102x，15x，求'1x，'2xd'1x'2x1x2xxd5.27'1x，55'2x55'1x，5.27'2x5.27'1x，33.18'2x33.18'1x，5.27'2x答案：A计算过程：5.2710155155211'1xxxxxx55515101510122'2xxxxxx4.将叉形（三叉）网络变成等值非叉形已知51x，102x，153x，20x，求'1x，'2x，'3xS2xSx1Sx3dxdSSS'1xx2'3x'33.83'1x，67.41'2x，125'3x67.41'1x，33.83'2x，125'3x125'1x，33.83'2x，67.41'3x11125'1x，67.41'2x，33.83'3x答案：B计算过程：67.4115101510205205323211'1xxxxxxxxx33.8315515520102010313122'2xxxxxxxxx12510510520152015212133'3xxxxxxxxx5.某变电所、简化接线如下dKukVA20000kV10.5d*TX=s∞=s∞%=10.5SS取MVASJ100,chK取1.8(1)求：变压器电抗标幺值TX*0.5B.0.525C.0.45D.0.55(2)求：短路点d的超瞬变短路电流周期分量有效值IA.kA48.10B.kA11C.kA22.12D.kA10(3)求：短路点d的短路全电流最大有效值chIA.kA93.13B.kA93.16C.kA93.14D.kA93.15(4)求：短路点d的短路冲击电流chiA.kA72.25B.kA72.27C.kA72.24D.kA72.26(5)求：10kV母线上的短路容量SA.MVA28.222B.MVA20012C.MVA6.190D.MVA86.181答案：(1)B,(2)A,(3)D,(4)D,(5)C计算过程：(1)525.0201001005.10100%*rjKTSSuX(2)kAUSIjjj5.55.1031003kAXIITj48.10525.05.5*(3)kAIIIKIchch93.1548.1052.152.118.12112122(4)kAIIIKichch72.2648.1055.255.228.12(5)MVAIUSj6.19048.105.10336.某变电所、简化接线如下kVA1600=300MVAskV10.5*SXddV400/230X*TKu%=4.5S取MVASJ100,chK取1.8(1)求：系统电抗标幺值s*XA.0.53B.0.23C．0.33D.0.43(2)求：变压器电抗标幺值T*XA.1.81B.3.81C.4.81D.2.81(3)求：短路点d的短路容量SA.MVA85.31B.MVA85.32C.MVA85.33D.MVA85.30(4)求：短路点d的超瞬变短路电流周期分量有效值IA.kA97.46B.kA97.43C.kA97.45D.kA97.44(5)求：d点短路时10kV侧的超瞬变穿越短路电流周期分量有效值dI（穿越）A.kA0.2B.kA75.113C.kA25.2D.kA5.1答案：(1)C,(2)D,(3)A,(4)C,(5)B计算过程：(1)33.0300100*SjSSSX(2)81.26.11001005.4100%*rjkTSSuX(3)MVAXXSSTSj85.3181.233.0100**(4)kAUSIj97.454.0385.313(5)kAId75.14.05.1097.45)(穿越7.某变电所有一台三绕组变压器，简化接线如下MVA45kV115kV6.3kV10.5*1XX*2X*3=9.4921Ku=16.431Ku=5.752Ku3123=s∞12dddd12S%%%取MVASJ100,chK取1.8(1)求：电阻值忽略不计时，变压器各级等值电抗标幺值1*X，2*X，3*X。224.01*X，141.02*X，013.03*X，013.01*X，224.02*X，141.03*X，013.01*X，141.02*X，224.03*X，224.01*X，013.02*X，141.03*X，(2)求：短路点1d的超瞬变短路电流周期分量有效值IA.kA07.25B.kA07.28C.kA07.26D.kA07.27(3)求：短路点1d的短路冲击电流chiA.kA47.46B.kA47.66C.kA47.56D.kA47.76(4)求：短路点2d的的超瞬变短路电流周期分量有效值IA.kA1.27B.kA1.26C.kA1.25D.kA1.2414(5)求：短路点2d的短路冲击电流chiA.kA99.63B.kA99.64C.kA99.62D.kA99.52答案：(1)D,(2)C,(3)B,(4)C,(5)A计算过程：(1)求：07.1075.54.1649.921%%%21%2313121KKKKuuuu58.04.1675.549.921%%%21%1323122KKKKuuuu33.649.975.54.1621%%%21%1223133KKKKuuuu224.04510010007.10100%11*rjkSSuX013.04510010058.0100%22*rjkSSuX141.04510010033.6100%33*rjkSSuX(2)求：kAUSIjjj5.55.1031003kAXXIIj07.26013.0224.05.52*1*(3)求：kAIIKichch47.6607.2655.228.12(4)求：kAUSIjjj16.93.631003kAXXIIj1.25141.0224.016.93*1*(5)求：kAIIKichch99.631.2555.228.128.请用标幺值进行下图各短路点的短路参数计算kV115=s∞架空线40kmLx=0.440115kVkV6.36.3kV31.5MVA=10.5ku%电缆0.5km=16ΩCx=0.080.5=0.04Ωd1d2kV6kA0.3=4%SXk取MVASj100已知各元件的电抗标幺值为：121.0*LX,333.0*TX,164.1*KX,1.0*CX*LX=0.121X=0.333*TX=1.164*KX=0.1*Cd2d1(1)求：1d点前的综合电抗标幺值A.0.333B.0.454C.0.121D.0.21215(2)求：2d前的综合电抗标幺值A.1.655B.1.676C.1.322D.1.718(3)求：1d点的超瞬变短路电流周期分量有效值IA.kA17.20B.kA17.19C.kA17.21D.kA17.22(4)求：2d点的超瞬变短路电流周期分量有效值IA.kA16.4B.kA16.3C.kA33.5D.kA16.6(5)求：2d点的短路容量SA.MVA39.45B.MVA2.58C.MVA22.67D.MVA48.34答案：(1)B,(2)D,(3)A,(4)C,(5)B计算过程：(1)求：454.0333.0121.0***TLXXX(2)求：718.11.0164.1454.0*****CKTLXXXXX(3)求：kAUSIjjj16.93.631003kAXXIITLj17.20454.016.9**(4)求：kAXXXXIICKTLj33.5718.116.9****(5)求：MVAIUSj2.5833.53.6339.请用有名值计算下图各短路点的短路参数kV115架空线40kmLX=0.440115kVkV6.36.3kV31.5MVA=10.5Ku%电缆0.5km=16ΩCX=0.080.5=0.04Ωd1d2kV6kA0.3=4%=s∞SXK取基准电压级为6.3kV已知各元件的电抗有名值并归算到6.3kV为：048.0LX,132.0TX,462.0KX,04.0CX2LX=0.048ΩX=0.132ΩTX=0.462ΩRX=0.04ΩCdd1(1)求：1d点前综合阻抗有名值A.048.0B.132.0C.18.0D.084.016(2)求：2d前的综合阻抗有名值A.665.0B.682.0C.657.0D.525.0(3)求：1d点的超瞬变短路电流周期分量有效值IA.kA2.18B.kA2.21C.kA2.20D.kA2.19(4)求：2d点的超瞬变短路电流周期分量有效值IA.kA16.6B.kA16.4C.kA16.3D.kA33.5(5)求：2d点的短路容量SA.MVA2.58B.kMVA39.45C.MVA48.34D.MVA2.67答案：(1)C,(2)B,(3)C,(4)D,(5)A计算过程：(1)求：18.0132.0048.0TLXXX(2)求：682.004.0462.018.0CKTLXXXXX(3)求：kAXXUITLj2.2018.033.63(4)求：kAXXXXUICKTLj33.5682.033.63(5)求：MVAIUSj2.5833.53.63310.某变电所主接线如下图所示d=2000MVAskV11510.5kV=10.5Ku%kV10kA2d250MVAS=8%XKX*K=0.209X=0.21*T=0.21*TX=0.05*SX取基准容量MVASj100,chK取1.8已知各元件的电抗标幺值为：05.0*SX,21.0*TX,209.0*KX(1)求：d点前的综合电抗标幺值A.0.19B.0.47C.0.26D.0.15517(2)求：d点的超瞬变短路电流周期分量有效值IA.kA8.25B.kA8.27C.kA8.26D.kA95.28(3)求：d点的短路全电流最大有效值chIA.kA21.39B.kA26.42C.kA44D.kA74.40(4)求：d点的短路冲击电流