EE 751

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 1

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 EE 751 Unsymmetrical Short Circuits 2

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 EE 751 Unsymmetrical Short Circuits 3

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 V 0 + V 1 = I 1 + V 2 = I 2 = I = 0. a /3 and Connect sequence networks in series at the fault (terminals), calculate I 0 , and I a = 3 I 0 E V 1 V 2 V 0 I 0 = I 1 = I 2 EE 751 Unsymmetrical Short Circuits 4

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% 1.0

j0.25

V 1 j0.25

V 2 j0.10

V 0 I 0 Use X'' d for both positive and negative sequence reactance I 0 = 1.0/(j0.60) = 1.667 /-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 EE 751 Unsymmetrical Short Circuits 5

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 I Then V 0 1 = V 2 = 0 and I 1 = (V a = -I 2 -V b )/3 = j I b /  3 E V 1 I 1 = -I 2 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 V 2 EE 751 Unsymmetrical Short Circuits 6

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% 1.0

j0.25

V 1 j0.25

V 2 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 I 1 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 EE 751 Unsymmetrical Short Circuits 7

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 and V b = V c = 0 = 0 E V 1 V 2 V 0 Then V 0 I 0 + I 1 = V + I 2 1 = 0 = V 2 = V a /3 I 1 I 2 I 0 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 EE 751 Unsymmetrical Short Circuits 8

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% 1.0

j0.25

V 1 I 1 j0.25

V 2 I 2 j0.10

V 0 I 0 I 1 = 1.0/(j0.321) = 3.11 /-90º V 1 = 1.0 – j0.25 I 1 = 0.222 /0º V 0 = V 2 = V 1 I 2 = 0.889 /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 EE 751 Unsymmetrical Short Circuits 9

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 V 0 = 0 and I b + V 1 + V 2 = I c = 0. Then I 0 = I 1 = I 2 = I a /3 and = 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.

EE 751 Unsymmetrical Short Circuits 10

G1 T1 L 1 f sc T2 G2 E EE 751 V 1 V 2 V 0 E E th Z 1th V 1 I 0 = I = I 1 2 Unsymmetrical Short Circuits Z 2th V 2 I 0 = I 1 = I 2 Z 0th V 0 11

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 T1: X = 0.08

= j 0.275, Z 0 = j 0.895

T2: X = 0.16

G1: X 1 G2: X 1 = X 2 = X 2 = 0.15 , X 0 = 0.50 , X 0 = 0.075 X n = 0.16

= 0.10

The circuit diagram shows the sequence networks connected to simulate the 1 f -ground fault EE 751 Unsymmetrical Short Circuits 12

EE 751 1.0

V 1 j0.15

j0.08

j0.275

j0.16

j0.50

V 2 j0.15

j0.08

j0.275

j0.16

j0.50

j0.30

V 0 j0.075

j0.08

j0.895 j0.16

j0.16

Unsymmetrical Short Circuits I 0 = I 1 = I 2 13

EE 751 1.0

V 1 j0.23

j0.935

V 2 j0.23

j0.935

V 0 j0.08

j1.055

Z Z Z th = Z 1th +Z 2th +Z 0th I 0 = 1/Z th I f 1th 0th = Z = 3 I 0 2th Unsymmetrical Short Circuits = j(0.23||0.935) = j0.1846

= j(0.08||1.055) = j0.0744

= j0.4435

= -j 2.255 per unit = -j 6.76 per unit 14

Use current division to find current from T1 in each sequence network: |I 0 | = 2.255

 1.055/(1.135) = 2.096

|I 1 | = |I 2 | = 2.255

 0.935/(1.165) = 1.810

And the transformation back to the line current from T1 gives: I a = -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. EE 751 Unsymmetrical Short Circuits 15

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 EE 751 Unsymmetrical Short Circuits 16

EE 751 b c a C A Winding connection for delta-wye transformer B V V V ca V bc bc ca V ab V CA V CN V AB V AN V BN V BC V BC V BN V AN V ab V CA V CN V AB Positive sequence +30º phase shift Negative sequence -30º phase shift Unsymmetrical Short Circuits 17

• 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 EE 751 Unsymmetrical Short Circuits 18

On the HV side of T1, the example gave: I 0 = -j 2.255

 1.055/(1.135) = -j 2.096 pu I 1 = I 2 = -j 2.255

 0.935/(1.165) = 1.810 /-90º pu On the LV side of T1: I 0 = 0 pu I 1 = 1.810 /-90º - 30º = 1.810 /-120º pu I 2 = 1.810 /-90º + 30º = 1.810 /-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 EE 751 Unsymmetrical Short Circuits 19

EE 751 I a b c a = 3.14 /-90º per unit C A B I A = 5.72/-90º per unit I B = I C = 0.29/-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).

Unsymmetrical Short Circuits 20

Open Conductor Faults

V a V b V c I a I b I c 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: EE 751 Unsymmetrical Short Circuits 21

0 0 V 0 I 0 1 1 V 1 I 1 2 2 V 2 I 2 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.

EE 751 Unsymmetrical Short Circuits 22

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 EE 751 Unsymmetrical Short Circuits 23

E EE 751 V V V G1 T1 1 2 0 L T2 1 f -gnd sc Motor Static loads ungrounded system E I 0 = 0 I a = 3I 0 = 0 The process can continue to operate until it can be shut down in an orderly fashion.

Unsymmetrical Short Circuits 24

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.

R 1th L 1th R 2th L 2th E V 1 E E C 0 V 2 V 0 I 0 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 EE 751 Unsymmetrical Short Circuits 25

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 T1 L T2 1 f -gnd sc Motor Static loads The process can continue to operate, but now transient voltages are damped much better and present less danger.

EE 751 Unsymmetrical Short Circuits 26

E V 1 E V 2 I 0 V 0 EE 751 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 27

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 EE 751 Unsymmetrical Short Circuits 28