Lecture 11: Unsymmetrical Short Circuits

Lecture 11: Unsymmetrical Short Circuits

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Lecture 11: Unsymmetrical Short Circuits

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1. Lecture 11: Unsymmetrical Short Circuits Symmetrical Components: Single-phase-to-ground short circuit Phase-to-phase short circuit Phase-to-phase-to-ground short circuit Unsymmetrical Short Circuits

2. Unbalanced Short Circuits • Procedure: • Set up all three sequence networks • Interconnect the networks at the point of the fault to simulate the short circuit • Calculate the 012 (sequence components) currents and voltages • Transform to ABC currents and voltages Unsymmetrical Short Circuits

3. Unbalanced Short Circuits • Short circuits considered: • single phase to ground = single line to ground • phase to phase = line to line short circuits • phase to phase to ground = double phase to ground = double line to ground = line to line to ground Unsymmetrical Short Circuits

4. Single-phase-to-ground short circuit on an unloaded generator Let phase a be the faulted phase: Va = 0 and Ib = Ic = 0. Then I0 = I1 = I2 = Ia/3 and V0 + V1 + V2 = 0. Connect sequence networks in series at the fault (terminals), calculate I0, and Ia = 3 I0 V1 E I0 = I1 = I2 V2 V0 Unsymmetrical Short Circuits

5. Example Wye-connected synchronous generator with neutral solidly grounded with single-phase-to-ground short circuit at terminals: Xd = 150%, X'd = 35%, X''d = 25%, X0 =10% Use X''d for both positive and negative sequence reactance I0 = 1.0/(j0.60) = 1.667 /-90º Ia = 3 I0 = 5.00 /-90º per unit Note that a 3-phase short circuit gives 4.00 per unit current, so most generators are not solidly grounded V1 1.0 j0.25 I0 V2 j0.25 V0 j0.10 Unsymmetrical Short Circuits

6. Phase-to-phase short circuit on an unloaded generator Let phase b be shorted to phase c: Ia = 0, Ic = -Ib and Vb = Vc Then V1 = V2 = (Va -Vb )/3I0 = 0 and I1 = -I2 = j Ib/3 V1 V2 E I1 = -I2 Connect the positive and negative sequence networks in parallel at the fault, calculate I1 Then Ib = -j3 I1 and Ic = 3 I1 Unsymmetrical Short Circuits

7. Example Wye-connected synchronous generator with phase-to-phase short circuit at terminals: Xd = 150%, X'd = 35%, X''d = 25%, X0 =10% I1 = -I2 = 1/j0.50 = 2.00 /-90º pu Ib = -j3 I1 = 3.46 /180 º pu Ic = j3 I1 = 3.46 /0 º pu V1 V2 1.0 j0.25 j0.25 I1 V1 = 1.0 – j 0.25 I1 = 0.50 /0º pu V2 = -j 0.25 I2 = 0.50 /0 º pu Va = 1.00 /0 º pu Vb = Vc = 0.50 /180 º pu Unsymmetrical Short Circuits

8. Phase-to-phase-to-ground short circuit on an unloaded generator Let phase b be shorted to phase c and also to ground: Ia = 0 and Vb = Vc = 0 Then V0 = V1 = V2 = Va/3 I0 + I1 + I2 = 0 V1 V2 V0 E I1 I2 I0 Connect all three sequence networks in parallel at the fault, calculate I0,I1 andI2. Then the symmetrical component transformation gives Ib and Ic Unsymmetrical Short Circuits

9. Example Wye-connected synchronous generator with neutral solidly grounded with phase-to-phase-to-ground short circuit at terminals: Xd = 150%, X'd = 35%, X''d = 25%, X0 =10% V1 V2 V0 1.0 j0.25 j0.25 j0.10 I1 I2 I0 I1 = 1.0/(j0.321) = 3.11 /-90ºV1 = 1.0 – j0.25 I1 = 0.222 /0º V0 = V2 = V1 I2 = 0.889 /90º I0 = 2.22 /90º Ia = 0.00 pu Ib = 4.81 /136.1º pu Ic = 4.81 /43.9º pu Unsymmetrical Short Circuits

10. Single-phase-to-ground short circuit on an unloaded power system Construct a Thevenin equivalent circuit for each sequence network. Let phase a be the faulted phase: Va = 0 and Ib = Ic = 0. Then I0 = I1 = I2 = Ia/3 and V0 + V1 + V2 = 0. Connect the sequence equivalents in series at the fault (terminals), calculate I0, and Ia = 3 I0. This current is total fault current. Line and other apparatus currents are found by solution of the sequence networks. Unsymmetrical Short Circuits

11. T1 T2 G1 L G2 1f sc V1 V1 E E Eth Z1th I0 = I1 = I2 I0 = I1 = I2 V2 V2 Z2th V0 V0 Z0th Unsymmetrical Short Circuits

12. G1: 100 MVA, 13.8 kV, X" = 15%, X0 = 7.5%, Xn = 10% G2: 50 MVA, 13.2 kV, X" = 25%, X0 = 8.0% T1: 100 MVA, 13.8 : 115 kV, X = 8.0% T2: 50 MVA, 13.2 : 115 kV, X = 8.0% Line: X1 = 36.4 ohms, X0 = 118 ohms Convert to per unit on 100 MVA base: Line impedance: Z1 = j 0.275, Z0 = j 0.895 T1: X = 0.08 T2: X = 0.16 G1: X1 = X2 = 0.15 , X0 = 0.075 Xn = 0.10 G2: X1 = X2 = 0.50 , X0 = 0.16 The circuit diagram shows the sequence networks connected to simulate the 1f-ground fault Unsymmetrical Short Circuits

13. 1.0 V1 j0.15 j0.08 j0.275 j0.16 j0.50 I0 = I1 = I2 V2 j0.15 j0.08 j0.275 j0.16 j0.50 V0 j0.30 j0.895 j0.16 j0.16 j0.075 j0.08 Unsymmetrical Short Circuits

14. 1.0 Z1th = Z2th = j(0.23||0.935) = j0.1846 V1 Z0th = j(0.08||1.055) = j0.0744 j0.23 j0.935 Zth = Z1th+Z2th+Z0th = j0.4435 V2 I0 = 1/Zth = -j 2.255 per unit If = 3 I0 = -j 6.76 per unit j0.23 j0.935 V0 j1.055 j0.08 Unsymmetrical Short Circuits

15. Use current division to find current from T1 in each sequence network: |I0| = 2.2551.055/(1.135) = 2.096 |I1| = |I2| = 2.2550.935/(1.165) = 1.810 And the transformation back to the line current from T1 gives: Ia = -j(1.810 + 1.810 + 2.096) = -j 5.72 per unit Note that on the LV side of T1, the zero-sequence line current is zero (due to the delta connection). The positive and negative sequence currents are shifted in phase by ±30 degrees, but in opposite directions. This is considered on the next slide. Unsymmetrical Short Circuits

16. Phase shifts in delta-wye transformers • Consider the delta-wye step-up transformer shown below: • The positive sequence shows a phase shift of 30º (hv side leading lv side) • The negative sequence has a phase shift of -30º C c A b a B Unsymmetrical Short Circuits

17. C c Winding connection for delta-wye transformer A b a B VCN VCA VAB Vca Positive sequence +30º phase shift VAN Vab Vab Vbc VBN VBC VBC Vbc VBN VAN Negative sequence -30º phase shift Vab VAB Vca VCN VCA Unsymmetrical Short Circuits

18. Consider the previous example, but now compute the currents at the generator G1 (the line currents on the low-voltage side of T1) The zero-sequence current is zero due to the transformer connection The LV side positive sequence current is shifted by –30º while the negative sequence current is shifted by +30º Unsymmetrical Short Circuits

19. On the HV side of T1, the example gave: I0 = -j 2.2551.055/(1.135) = -j 2.096 pu I1 = I2 = -j 2.2550.935/(1.165) = 1.810 /-90º pu On the LV side of T1: I0 = 0 pu I1 = 1.810 /-90º - 30º = 1.810 /-120º pu I2 = 1.810 /-90º + 30º = 1.810 /-60º pu Ia = I0+I1+I2 = 3.14 /-90º per unit Ib = I0+a2I1+aI2 = 3.14 /90º per unit Ic = I0+aI1+a2I2 = 0.00 per unit Unsymmetrical Short Circuits

20. C c A b a B Ia = 3.14 /-90º per unit IA = 5.72/-90º per unit IB = IC = 0.29/-90º per unit Note that IB and IC create small circulating currents in the delta side of both transformers (only one of which is shown). Unsymmetrical Short Circuits

21. Open Conductor Faults Ia Va Vb Ib Vc Ic Single open conductor in a line: Vb = Vc = 0 Ia = 0 I0 = (Ib + Ic)/3 I1 = (a Ib + a2 Ic)/3 I2 = (a2 Ib + a Ic)/3 I0 + I1 + I2 = 0 V0 = V1 = V2 = Va/3 So connect the three sequence networks in parallel at the point of the open circuit as shown below: Unsymmetrical Short Circuits

22. 0 0 V0 I0 1 1 V1 I1 2 2 V2 I2 Notice that positive sequence currents see an impedance that is on the order of twice the normal value. Power is transferred, but with the creation of negative sequence currents. Unsymmetrical Short Circuits

23. Ungrounded Power Delivery Systems • Many process industries have trouble with unplanned process shut-down due to faults • Since the most common fault is a short circuit from single phase to ground, why not use an ungrounded system? • Supply power from a delta-delta or wye-delta step-down transformer and the LV system is ungrounded Unsymmetrical Short Circuits

24. T1 T2 G1 1f-gnd sc L Motor Static loads ungrounded system V1 E E I0 = 0 V2 Ia = 3I0 = 0 The process can continue to operate until it can be shut down in an orderly fashion. V0 Unsymmetrical Short Circuits

25. But include stray capacitance to ground, and there is an RLC series circuit that can produce high-frequency transients that may be lightly damped. This can cause problems with transient overvoltages. R1th L1th R2th L2th V1 E E C0 E I0 If the fault is a repetitive, arcing short-circuit, large transient voltages to ground can be produced. This may damage insulation and lead to burn-down of the system V2 V0 Unsymmetrical Short Circuits

26. The solution is to use high-resistance grounding to limit the single phase to ground short circuit current to a very small value, but greater than the charging current. T1 T2 G1 1f-gnd sc L Motor Static loads The process can continue to operate, but now transient voltages are damped much better and present less danger. Unsymmetrical Short Circuits

27. V1 E E I0 V2 V0 Now every part of the system is grounded and the plant step-down transformer provides high-resistance grounding to its distribution circuits. Unsymmetrical Short Circuits

28. Discussion • Many other faults have been analyzed using symmetrical components • See, for example, Electrical Transmission and Distribution Reference Book, Westinghouse Electric Corporation, 1964 • Another approach for more complicated cases: use three-phase primitive branch impedance matrix and transform to three-phase bus admittance matrix in abc coordinates Unsymmetrical Short Circuits