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EE 751Unsymmetrical Short Circuits1 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

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EE 751Unsymmetrical Short Circuits2 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

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EE 751Unsymmetrical Short Circuits3 Unbalanced Short Circuits Short circuits considered: 1.single phase to ground = single line to ground 2.phase to phase = line to line short circuits 3.phase to phase to ground = double phase to ground = double line to ground = line to line to ground

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EE 751Unsymmetrical Short Circuits4 Single-phase-to-ground short circuit on an unloaded generator Let phase a be the faulted phase: V a = 0 and I b = I c = 0. Then I 0 = I 1 = I 2 = I a /3 and V 0 + V 1 + V 2 = 0. Connect sequence networks in series at the fault (terminals), calculate I 0, and I a = 3 I 0 I 0 = I 1 = I 2 V1V1 V2V2 V0V0 E

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EE 751Unsymmetrical Short Circuits5 Example Wye-connected synchronous generator with neutral solidly grounded with single-phase-to-ground short circuit at terminals: X d = 150%, X' d = 35%, X'' d = 25%, X 0 =10% I 0 V1V1 V2V2 V0V0 1.0 Use X'' d for both positive and negative sequence reactance I 0 = 1.0/(j0.60) = /-90º I a = 3 I 0 = 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 j0.25 j0.10

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EE 751Unsymmetrical Short Circuits6 Phase-to-phase short circuit on an unloaded generator Let phase b be shorted to phase c: I a = 0, I c = -I b and V b = V c Then V 1 = V 2 = (V a -V b )/3 I 0 = 0 and I 1 = -I 2 = j I b / 3 I 1 = -I 2 V1V1 V2V2 E Connect the positive and negative sequence networks in parallel at the fault, calculate I 1 Then I b = -j 3 I 1 and I c = 3 I 1

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EE 751Unsymmetrical Short Circuits7 Example Wye-connected synchronous generator with phase- to-phase short circuit at terminals: X d = 150%, X' d = 35%, X'' d = 25%, X 0 =10% V1V1 V2V2 1.0 I 1 = -I 2 = 1/j0.50 = 2.00 /-90º pu I b = -j 3 I 1 = 3.46 /180 º pu I c = j 3 I 1 = 3.46 /0 º pu j0.25 I1I1 V 1 = 1.0 – j 0.25 I 1 = 0.50 /0º pu V 2 = -j 0.25 I 2 = 0.50 /0 º pu V a = 1.00 /0 º pu V b = V c = 0.50 /180 º pu

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EE 751Unsymmetrical Short Circuits8 Phase-to-phase-to-ground short circuit on an unloaded generator Let phase b be shorted to phase c and also to ground: I a = 0 and V b = V c = 0 Then V 0 = V 1 = V 2 = V a /3 I 0 + I 1 + I 2 = 0 I1I1 V1V1 V2V2 E Connect all three sequence networks in parallel at the fault, calculate I 0, I 1 and I 2. Then the symmetrical component transformation gives I b and I c V0V0 I2I2 I0I0

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EE 751Unsymmetrical Short Circuits9 Example Wye-connected synchronous generator with neutral solidly grounded with phase-to-phase-to-ground short circuit at terminals: X d = 150%, X' d = 35%, X'' d = 25%, X 0 =10% V1V1 V2V2 V0V0 1.0 I 1 = 1.0/(j0.321) = 3.11 /-90º V 1 = 1.0 – j0.25 I 1 = /0º V 0 = V 2 = V 1 I 2 = /90º I 0 = 2.22 /90º I a = 0.00 pu I b = 4.81 /136.1º pu I c = 4.81 /43.9º pu j0.25 j0.10 I 1 I 2 I 0

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EE 751Unsymmetrical Short Circuits10 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: V a = 0 and I b = I c = 0. Then I 0 = I 1 = I 2 = I a /3 and V 0 + V 1 + V 2 = 0. Connect the sequence equivalents in series at the fault (terminals), calculate I 0, and I a = 3 I 0. This current is total fault current. Line and other apparatus currents are found by solution of the sequence networks.

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EE 751Unsymmetrical Short Circuits11 I 0 = I 1 = I 2 V1V1 V2V2 V0V0 E th Z 1th Z 2th Z 0th V1V1 V2V2 V0V0 EE I 0 = I 1 = I 2 1 sc G1 G2 T2T1 L

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EE 751Unsymmetrical Short Circuits12 G1: 100 MVA, 13.8 kV, X" = 15%, X 0 = 7.5%, X n = 10% G2: 50 MVA, 13.2 kV, X" = 25%, X 0 = 8.0% T1: 100 MVA, 13.8 : 115 kV, X = 8.0% T2: 50 MVA, 13.2 : 115 kV, X = 8.0% Line: X 1 = 36.4 ohms, X 0 = 118 ohms Convert to per unit on 100 MVA base: Line impedance: Z 1 = j 0.275, Z 0 = j T1: X = 0.08T2: X = 0.16 G1: X 1 = X 2 = 0.15, X 0 = X n = 0.10 G2: X 1 = X 2 = 0.50, X 0 = 0.16 The circuit diagram shows the sequence networks connected to simulate the 1 -ground fault

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EE 751Unsymmetrical Short Circuits13 V1V1 V2V2 V0V0 1.0 I 0 = I 1 = I 2 j0.16 j0.15 j0.075 j0.30 j0.275 j0.08 j0.275 j0.895j0.16 j0.50 j0.16 j0.50

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EE 751Unsymmetrical Short Circuits14 V1V1 V2V2 V0V0 1.0 I 0 = 1/Z th = -j per unit j0.23 j0.935 j0.08 j0.935 j1.055 Z 1th = Z 2th = j(0.23||0.935) = j Z 0th = j(0.08||1.055) = j Z th = Z 1th +Z 2th +Z 0th = j I f = 3 I 0 = -j 6.76 per unit

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EE 751Unsymmetrical Short Circuits15 Use current division to find current from T1 in each sequence network: |I 0 | = /(1.135) = |I 1 | = |I 2 | = /(1.165) = And the transformation back to the line current from T1 gives: I a = -j( ) = -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.

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EE 751Unsymmetrical Short Circuits16 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 A a B b c C

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EE 751Unsymmetrical Short Circuits17 V ab V AB V BC V bc V ab V AB V BC V bc V CA V ca A a B b c C V CA V ab V AN V BN V CN V AN V BN V CN Positive sequence +30º phase shift Negative sequence -30º phase shift Winding connection for delta-wye transformer

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EE 751Unsymmetrical Short Circuits18 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

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EE 751Unsymmetrical Short Circuits19 On the HV side of T1, the example gave: I 0 = -j /(1.135) = -j pu I 1 = I 2 = -j /(1.165) = /-90º pu On the LV side of T1: I 0 = 0 pu I 1 = /-90º - 30º = /-120º pu I 2 = /-90º + 30º = /-60º pu I a = I 0 +I 1 +I 2 = 3.14 /-90º per unit I b = I 0 +a 2 I 1 +aI 2 = 3.14 /90º per unit I c = I 0 +aI 1 +a 2 I 2 = 0.00 per unit

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EE 751Unsymmetrical Short Circuits20 A a B b c C I a = 3.14 /-90º per unit I A = 5.72/-90º per unit Note that I B and I C create small circulating currents in the delta side of both transformers (only one of which is shown). I B = I C = 0.29/-90º per unit

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EE 751Unsymmetrical Short Circuits21 Open Conductor Faults VaVa VbVb VcVc IaIa IbIb IcIc Single open conductor in a line: V b = V c = 0 I a = 0 I 0 = (I b + I c )/3 I 1 = (a I b + a 2 I c )/3 I 2 = (a 2 I b + a I c )/3 I 0 + I 1 + I 2 = 0 V 0 = V 1 = V 2 = V a /3 So connect the three sequence networks in parallel at the point of the open circuit as shown below:

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EE 751Unsymmetrical Short Circuits22 V0V0 00 I0I0 V1V1 11 I1I1 V2V2 22 I2I2 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.

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EE 751Unsymmetrical Short Circuits23 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

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EE 751Unsymmetrical Short Circuits24 V1V1 V2V2 V0V0 EE I 0 = 0 G1 Motor T2T1 L Static loads ungrounded system 1 -gnd sc I a = 3I 0 = 0 The process can continue to operate until it can be shut down in an orderly fashion.

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EE 751Unsymmetrical Short Circuits25 EE I 0 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. E R 1th L 1th R 2th L 2th C0C0 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 V1V1 V2V2 V0V0

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EE 751Unsymmetrical Short Circuits26 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. G1 Motor T2T1 L Static loads 1 -gnd sc The process can continue to operate, but now transient voltages are damped much better and present less danger.

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EE 751Unsymmetrical Short Circuits27 EE V1V1 V2V2 V0V0 Now every part of the system is grounded and the plant step-down transformer provides high-resistance grounding to its distribution circuits. I 0

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EE 751Unsymmetrical Short Circuits28 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

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