453 Appendix A: Fault Calculations In this appendix, asymmetrical fault calculations for a transformer are given. A single-line-to-ground fault on LV side is considered. The neutrals of LV and HV windings are solidly grounded. Two distinct cases are considered, viz. with and without fault in-feed from the secondary (LV) side. For each of these two cases, two conditions are considered, viz. with and without stabilizing tertiary winding (TV). The effects of omitting tertiary winding are illustrated with these cases. Transformer Data: Rating (S)=25 MVA, Core construction: three-phase three-limb Voltage ratio: HV/LV/TV, 132/11/11 KV, star/star/delta Leakage sequence impedances (based on 25 MVA base): HV/LV:Z 1 =Z 2 =12%, Z 0 = 10.2% HV/TV:Z 1 =Z 2 =20%, Z 0 =17.0% LV/TV: Z 1 =Z 2 =6%, Z 0 =5.4% 132 KV system fault level (S F )=5000 MVA The positive-sequence system impedance on the HV side is Z 1HS =0.5% (calculated by equation 6.2). The HV and LV side systems are assumed as effectively grounded systems (X 0 /X 1 not greater than 3). Here, in particular, the zero-sequence system impedances on the LV and HV sides are assumed to be equal to the corresponding positive-sequence system impedances for simplicity. As mentioned earlier, two alternatives are considered for 11 KV (LV) system: (i) Without any fault in-feed from 11 KV system. (ii) With fault in-feed from 11 KV system: Copyright © 2004 by Marcel Dekker, Inc. Appendix A454 11 KV system fault level (S F )=833.33 MVA, which gives the positive- sequence system impedance on LV side: Z 1LS =3.0% (by equation 6.2). The equivalent impedances in the star equivalent circuit can be computed by using the theory given in Section 3.5 (Chapter 3) as ZH=(Z HL +Z HT -Z LT )/2 Z L =(Z HL +Z LT -Z HT )/2 Z T =(Z HT +Z LT -Z HL )/2 Therefore, the corresponding positive-, negative- and zero-sequence impedances of HV, LV and TV windings are computed as follows: Z 1H =Z 2H =(12+20-6)/2=l3%, Z 0H =(10.2+17-5.4)/2=10.9% Z 1T =Z 2T =(20+6-12)/2=7%, Z 0T =(17+5.4-10.2)/2= 6.1% A1 Asymmetrical Fault with No In-Feed from LV Side A1.1 Delta connected tertiary winding is present The equivalent network under the fault condition is shown in figure A1 for a single-line-to-ground fault on the LV side. All the impedances are in % values. taken to be effectively infinite since there is no in-feed from the LV side. Figure A1 Sequence network for Case A1.1 Z 1 L=Z 2L =(12+6-20)/2=-l%, Z 0L= (10.2+5.4-17)/2 =-0.7% The network can be derived from figure 6.3 (a) with impedances Z 1LS , Z 2LS and Z 0LS Copyright © 2004 by Marcel Dekker, Inc. Appendix A 455 Total impedance=2(0.5+13–1)-0.7+(11.4//6.1)=28.27% For a single phase-earth fault, using equations 6.6 and 6.7 we get the values of sequence components of currents as I 1 =I 2 =I 0 =100/28.27=3.54 p.u. (i.e., 3.54 times the rated current). Total fault current=I 1 +I 2 +I 0 =3(3.54)=10.62 p.u. The fault current in any of the windings is calculated by adding the corresponding sequence currents flowing in them in the three sequence networks. The neutral of LV and HV windings carries 3 times the corresponding zero-sequence current flowing in them. Therefore, Current through HV winding=3.54+3.54+3.54 (6.1/17.5)=8.31 p.u. Current through LV winding=3.54+3.54+3.54=10.62 p.u. Current through stabilizing winding=3.54 (11.4/17.5)=2.31 p.u. Current through HV neutral=3×3.54 (6.1/17.5)=3.7 p.u. Current through LV neutral=total fault current=3×3.54 =10.62 p.u. A1.2 Delta connected tertiary winding is absent The equivalent network under the fault condition is shown in figure A2 for a single-line-to-ground fault on the LV side. Proceeding in the similar way as in Section A1.1 we get Total Impedance=2(0.5+13–1)-0.7+0.5+10.9=35.7% Figure A2 Sequence network for Case A1.2 Copyright © 2004 by Marcel Dekker, Inc. Appendix A456 I 1 =I 2 =I 0 =100/35.7=2.8 p.u. Total fault current=I 1 +I 2 +I 0 =3 (2.8)=8.4 p.u. Current through HV winding=2.8+2.8+2.8=8.4 p.u. Current through LV winding=2.8+2.8+2.8=8.4 p.u. Current through HV neutral=8.4 p.u. Current through LV neutral=8.4 p.u. It is to be noted that the single-line-to-ground fault current in this case (8.4 p.u.) is more than the three-phase fault current (i.e., 1/Z 1 =1/0.12=8.33 p.u.) due to two reasons. First, the zero-sequence leakage impedance of the transformer is less than the positive-sequence leakage impedance. Second, it is assumed that the positive- and zero-sequence system impedances of HV side are equal (0.5 p.u.). In many of the cases, the system zero-sequence impedance is more than the positive- sequence impedance, which will make the single-line-to-ground fault current lower than the three-phase fault current. It can be observed from the above two case studies that the absence of the tertiary stabilizing winding reduces the total fault current and the current in the faulted winding (LV). However, the current in HV neutral increases significantly. Figure A3 Sequence network for Case A2.1 Copyright © 2004 by Marcel Dekker, Inc. Appendix A 457 A2 Asymmetrical Fault with In-Feed from LV Side A2.1 Delta connected tertiary winding is present The equivalent network under the fault condition is shown in figure A3 for a single-line-to-ground fault on the LV side. The network is same as that given in figure 6.3. Total positive-sequence impedance=12.5//3=2.42% Total negative-sequence impedance=12.5//3=2.42% Total zero-sequence impedance=[(11.4//6.1)-0.7]//3=1.56% Total impedance=2.42+2.42+1.56=6.4% I 1 =I 2 =I 0 =100/6.4=15.62 p.u. Total fault current=I 1 +I 2 +I 0 =3 (15.62)=46.86 p.u. Current through HV winding =15.62(3/15.5)+15.62(3/15.5)+15.62(3/6.27)(6.1/17.5) =3.02+3.02+2.6=8.64 p.u. Current through LV winding=3.02+3.02+15.62(3/6.27)=13.51 p.u. Current through stabilizing winding=15.62(3/6.27)(11.4/17.5)=4.87 p.u. Current through HV neutral=3×2.6=7.8 p.u. Current through LV neutral=3×7.47=22.41 p.u. Figure A4 Sequence network for Case A2.2 Copyright © 2004 by Marcel Dekker, Inc. Appendix A458 A2.2 Delta connected tertiary winding is absent The equivalent network under the fault condition is shown in figure A4 for a single-line-to-ground fault on the LV side. Proceeding in the similar way as in Section A2.1 we get Total positive-sequence impedance=12.5//3=2.42% Total negative-sequence impedance=12.5//3=2.42% Total zero-sequence impedance=10.7//3=2.34% Total Impedance=2.42+2.42+2.34–7.18% I 1 =I 2 =I 0 =100/7.18 7.18=13.93 p.u. Total fault current=I 1 +I 2 +I 0 =3(13.93)=41.79 p.u. Current through HV winding =13.93(3/15.5)+13.93(3/15.5)+13.93(3/13.7) =2.7+2.7+3.05=8.45 p.u. Current through LV winding=2.7+2.7+3.05=8.45 p.u. Current through HV neutral=3×3.05=9.15 p.u. Current through LV neutral=3×3.05=9.15 p.u. It can be observed from the above two case studies (in-feed from LV side) that the absence of the tertiary stabilizing winding reduces the total fault current and the current in the faulted winding (LV). The current in HV neutral increases. Copyright © 2004 by Marcel Dekker, Inc. . Impedance=2.42+2.42+2.34–7 .18 % I 1 =I 2 =I 0 =10 0/7 .18 7 .18 =13 .93 p.u. Total fault current=I 1 +I 2 +I 0 =3 (13 .93)= 41. 79 p.u. Current through HV winding =13 .93(3 /15 .5) +13 .93(3 /15 .5) +13 .93(3 /13 .7) =2.7+2.7+3.05=8.45. HV/LV/TV, 13 2 /11 /11 KV, star/star/delta Leakage sequence impedances (based on 25 MVA base): HV/LV:Z 1 =Z 2 =12 %, Z 0 = 10 .2% HV/TV:Z 1 =Z 2 =20%, Z 0 =17 .0% LV/TV: Z 1 =Z 2 =6%, Z 0 =5.4% 13 2 KV. negative- and zero-sequence impedances of HV, LV and TV windings are computed as follows: Z 1H =Z 2H = (12 +20-6)/2=l3%, Z 0H = (10 .2 +17 -5.4)/2 =10 .9% Z 1T =Z 2T =(20+6 -12 )/2=7%, Z 0T = (17 +5.4 -10 .2)/2=