Lecture 11: Unsymmetrical Short Circuits

EE 751 Unsymmetrical Short Circuits 1

Lecture 11: Unsymmetrical Short

CircuitsSymmetrical Components:

Single-phase-to-ground short circuit

Phase-to-phase short circuitPhase-to-phase-to-ground short

circuit


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


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


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

I0

= I1

= I2

V1

V2

V0

E


ExampleWye-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%

I0

V1

V2

V0

1.0Use 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

j0.25

j0.25

j0.10


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 I1 = -I2

V1 V2E

Connect the positive and negative sequence networks in parallel at the fault, calculate I1

Then Ib = -j3 I1 and Ic = 3 I1


ExampleWye-connected synchronous generator with phase-to-phase short circuit at terminals: Xd = 150%, X'd = 35%, X''d = 25%, X0 =10%

V1 V21.0I1 = -I2 = 1/j0.50 = 2.00 /-90º pu

Ib = -j3 I1 = 3.46 /180 º pu

Ic = j3 I1 = 3.46 /0 º pu

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


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

I1

V1 V2E

Connect all three sequence networks in parallel at the fault, calculate I0, I1 and I2. Then the symmetrical component transformation gives Ib and Ic

V0

I2 I0


ExampleWye-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 V01.0

I1 = 1.0/(j0.321) = 3.11 /-90º V1 = 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

j0.25 j0.25 j0.10 I1 I2 I0


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.


I0

= I1

= I2

V1

V2

V0

Eth Z1th

Z2th

Z0th

V1

V2

V0

E E

I0

= I1

= I2

1 sc

G1 G2T2T1 L


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 0.895T1: X = 0.08 T2: X = 0.16G1: X1 = X2 = 0.15 , X0 = 0.075 Xn = 0.10G2: X1 = X2 = 0.50 , X0 = 0.16

The circuit diagram shows the sequence networks connected to simulate the 1-ground fault


V1

V2

V0

1.0

I0

= I1

= I2j0.16

j0.15

j0.15

j0.075

j0.30

j0.275

j0.08

j0.08

j0.08

j0.275

j0.895 j0.16

j0.16

j0.50

j0.16

j0.50


V1

V2

V0

1.0

I0 = 1/Zth = -j 2.255 per unit

j0.23

j0.23

j0.935

j0.08

j0.935

j1.055

Z1th = Z2th = j(0.23||0.935) = j0.1846

Z0th = j(0.08||1.055) = j0.0744

Zth = Z1th+Z2th+Z0th = j0.4435

If = 3 I0 = -j 6.76 per unit


Use current division to find current from T1 in each sequence network:|I0| = 2.2551.055/(1.135) = 2.096|I1| = |I2| = 2.2550.935/(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.


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

Aa

Bb

c C


Vab

VAB

VBC

Vbc

Vab VAB

VBC

Vbc

VCA

Vca

Aa

Bb

c C

Vca

VCA

VabVAN

VBN

VCN

VAN

VBN

VCN

Positive sequence +30º phase shift

Negative sequence -30º phase shift

Winding connection for delta-wye transformer


• 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 connectionThe LV side positive sequence current is

shifted by –30º while the negative sequence current is shifted by +30º


On the HV side of T1, the example gave:

I0 = -j 2.2551.055/(1.135) = -j 2.096 pu

I1 = I2 = -j 2.2550.935/(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


Aa

Bb

cC

Ia = 3.14 /-90º

per unit

IA = 5.72/-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).

IB = IC = 0.29/-90º

per unit


Open Conductor FaultsVa

Vb

Vc

Ia

Ib

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:


V0

00

I0

V1

11

I1

V2

22

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.


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


V1

V2

V0

E E

I0 = 0

G1Motor

T2T1 L

Static loadsungrounded system

1-gnd sc

Ia = 3I0 = 0 The process can continue to operate until it can be shut down in an orderly fashion.


E E

I0

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

R1th L1th R2th L2th

C0

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

V1

V2

V0


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.

G1Motor

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.


E EV1

V2

V0

Now every part of the system is grounded and the plant step-down transformer provides high-resistance grounding to its distribution circuits.

I0


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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Lecture 11: Unsymmetrical Short Circuits