Grounding Examples. MVA 1000kW. pu 1 A a 2

Transcription

1 Session 2; Page 1/7 Fall 216 Grounding Examples a 1e j12deg MVA 1kW pu 1 A 12 1 a 2 a 1 a a 2 A 416 V feeder is supplied by the WYE connected side of a 75 MVA transformer. The system MVA sc supplying the delta side of the transformer is 65 MVA. The transformer has a leakage reactance of 1%. A ground impedance will be connected in the neutral of 4.16kV side of the transformer to limit fault currents. A Sketch the per unit diagram for the system 1MVA Srated 75MVA V_LL 4.16kV V_LL V_ln V_ln 2.42kV MVAsc 65MVA 1. 2 _pu _pu.154pu MVAsc X_xfmr.1 X_xfmr 416V 416V.1pu 2 Srated V src j.154pu j.1pu

2 Session 2; Page 2/7 Fall 216 B Determine sequence networks for the system Positive Sequence: Negative Sequence: j.154pu j.1pu j.154pu j.1 pu V src Zero Sequence: j.154pu j.1pu Z gnd Note that this is a -Y grounded transformer. Also, assuming that zero sequence leakage impedance equal to positive and negative sequence values. C Assume that the feeder is all overhead lines with negligable capacitance. Determine the ground reactance needed to limit the single line to ground fault current to 6A. If_slgmax 6A Ibase Ibase 1.88kA V_LL If_slgmax Ifpu Ifpu.42pu Ibase Zbase V_LL 2 For a SLG fault we have (connect positive, negative and zero sequence circuits in series): I = Z1 Z2 Z j Xgnd where 1. e j9 deg Ifpu and we know for a SLG fault: I Solve for Zgnd Z1 j_pu jx_xfmr Z2 Z1 Z jx_xfmr 1 Zgnd ( Z1 Z2 Z) I Zgnd 2.772i per unit

3 Session 2; Page /7 Fall 216 Xgndpu Im( Zgnd) Xgndpu 2.77 per unit Xgnd XgndpuZbase Xgnd.59 Ω Xgnd Lgnd 2π6Hz Lgnd.954 mh at 6Hz D If the feeder is largely underground, the capacitance cannot be neglected. If the total per phase capacitance to ground is 1.5 F, determine the grounding resistance needed to limit the single line to ground fault current to 2 A. Cparasitic 1.5μF 1 Xc 2π6Hz Cparasitic Xc kω Xc Xc_pu Xc_pu pu Zbase Islg_max 2A Islg_max Islgpu Islgpu pu Ibase The sequence networks will now change with the addition of the capacitance as shown. j.154 pu j.1pu Positive Sequence -j1218.6pu j.154 pu j.1pu I Negative Sequence -j1218.6pu j.154 pu j.1pu Zero Sequence R gnd -j1218.6pu

4 Session 2; Page 4/7 Fall 216 I = Z1 Z2 [( Z Rg) ( jxc) ] Z Rg jxc Note that Z1+Z2 will be much much smaller than the parallel combination of R and -jxc, to that Z1 and Z2 can be neglected, as can Z We also, only care about the magnitude of the reduced current, not the angle. So we are actually solving: I = [( ZRg) ( jxc) ] ZRgjXc which requires an iterative solution. I Islgpu Initial Guess: Rg 1 MathCAD solve block: Given I = ( Rg Z) ( jxc_pu) ( Rg Z) jxc_pu Rgnd_pu Find( Rg) Rgnd_pu Rgnd Rgnd_puZbase Rgnd Ω

5 Session 2; Page 5/7 Fall 216 E Calculate the line to ground voltages on the unfaulted phases in parts C and D and calculate the zero sequence voltages and currents. Part C: I_partC Z jxgndpu Z1 Z2 I_partC.144 I1_partC I_partC I2_partC I_partC as a check: Ia_partC I_partC Ia_partC.42 pu Ia_partC Ibase 6kA V1_partC I1_partCZ1 V1_partC.959i V2_partC I2_partCZ2 V2_partC.41i V_partC I_partC( Z jxgndpu) V_partC.917i V_partC Vabc_partC A 12 V1_partC V2_partC Vabc_partC i i Vabc_partC arg( Vabc_partC) deg V_ln Vabc_partC kv Part D: ZgndD ( Rgnd_pu Z) ( jxc_pu) ( Rgnd_pu Z) jxc_pu I_partD ZgndD Z1 Z2 I_partD i 1 4 I1_partD I_partD I2_partD I_partD I_partD

6 Session 2; Page 6/7 Fall 216 as a check: Ia_partD I_partD Ia_partD pu Ia_partD Ibase 2.1A V1_partD I1_partDZ1 V1_partD 1 V2_partD I2_partDZ2 V2_partD V_partD I_partD( ZgndD) V_partD 1 V_partD Vabc_partD A 12 V1_partD V2_partD Vabc_partD i i Vabc_partD V_ln Vabc_partD kv arg( Vabc_partD) deg F Compute the single line to ground fault current and the voltage on the unfaulted phases if the transformer is solidly grounded. Calculate the zero sequence voltages and currents. I_gnd Z Z1 Z2 I_gnd 1.41 I1_gnd I_gnd I2_gnd I_gnd Ia_gnd I_gnd Ia_gnd 4.29 per unit Ia_gndIbase 58.8kA V1_gnd I1_gndZ1 V1_gnd.594i V2_gnd I2_gndZ2 V2_gnd.46i V_gnd I_gndZ V_gnd.18841i

7 Session 2; Page 7/7 Fall 216 V_gnd 1 16 Vabc A 12 V1_gnd V2_gnd Vabc i i Vabc arg( Vabc) deg Note that Vb and Vc are nearly 1. per unit, and are slightly depressed. If Z=Z1=Z2 then they would be 1. and offset from each other by 12 degrees. V_LL Vabc kv G For the different grounded cases described above, discuss the available quantities to measure for ground fault protection and suggest a scheme to consider (based on voltage, current, etc). For the high resistance grounded case, there isn't enough current available for doing ground fault protection, but V is 1pu, so there is enough voltage available to use that to identify the presense of a ground fault. For the case with the low inductance grounded, there is sufficient I for detecting a fault, although any fault impedance (resistance) may make this too difficult. There is also probably sufficient V to use that to detect the fault. For the solidly grounded case, V is pretty small, and it might be hard to discriminate sufficiently to identify a fault. On the other hand, I is large.