EE 751Unsymmetrical Short Circuits1 Lecture 11: Unsymmetrical Short Circuits Symmetrical Components:...
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Transcript of EE 751Unsymmetrical Short Circuits1 Lecture 11: Unsymmetrical Short Circuits Symmetrical Components:...
EE 751 Unsymmetrical Short Circuits 1
Lecture 11: Unsymmetrical Short
CircuitsSymmetrical Components:
Single-phase-to-ground short circuit
Phase-to-phase short circuit
Phase-to-phase-to-ground short circuit
EE 751 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
EE 751 Unsymmetrical Short Circuits 3
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 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
I0
= I1
= I2
V1
V2
V0
E
EE 751 Unsymmetrical Short Circuits 5
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
EE 751 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 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
EE 751 Unsymmetrical Short Circuits 7
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
EE 751 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
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
EE 751 Unsymmetrical Short Circuits 9
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
EE 751 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.
EE 751 Unsymmetrical Short Circuits 11
I0
= I1
= I2
V1
V2
V0
Eth Z1th
Z2th
Z0th
V1
V2
V0
E E
I0
= I1
= I2
1 sc
G1 G2T2T1 L
EE 751 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 0.895T1: 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
EE 751 Unsymmetrical Short Circuits 13
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
EE 751 Unsymmetrical Short Circuits 14
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
EE 751 Unsymmetrical Short Circuits 15
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.
EE 751 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
Aa
Bb
c C
EE 751 Unsymmetrical Short Circuits 17
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
EE 751 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
EE 751 Unsymmetrical Short Circuits 19
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
EE 751 Unsymmetrical Short Circuits 20
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
EE 751 Unsymmetrical Short Circuits 21
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:
EE 751 Unsymmetrical Short Circuits 22
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.
EE 751 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
EE 751 Unsymmetrical Short Circuits 24
V1
V2
V0
E E
I0 = 0
G1Motor
T2T1 L
Static loads
ungrounded system
1-gnd sc
Ia = 3I0 = 0
The process can continue to operate until it can be shut down in an orderly fashion.
EE 751 Unsymmetrical Short Circuits 25
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
EE 751 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.
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.
EE 751 Unsymmetrical Short Circuits 27
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
EE 751 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