SHORT CIRCUIT CALCULATIONS REVISITED FORTUNATO C. LEYNES Chairman Professional Regulatory Board of Electrical Engineering Professional Regulation Commission Short Circuit Calculations IIEE Presentation
Short Circuit (Fault) Analysis FAULTPROOF SYSTEM not practical neither economical faults or failures occur in any power system
In the various parts of the electrical network under short circuit or unbalanced condition, the determination of the magnitudes and phase angles Currents Voltages Impedances Short Circuit Calculations IIEE Presentation
Application of Fault Analysis 1. The determination of the required mechanical strength of electrical equipment to withstand the stresses brought about by the flow of high short circuit currents 2. The selection of circuit breakers and switch ratings 3. The selection of protective relay and fuse ratings
Short Circuit Calculations IIEE Presentation
Application of Fault Analysis 4. The setting and coordination of protective devices 5. The selection of surge arresters and insulation ratings of electrical equipment 6. The determination of fault impedances for use in stability studies 7. The calculation of voltage sags caused resulting from short circuits
Short Circuit Calculations IIEE Presentation
Application of Fault Analysis 8. The sizing of series reactors to limit the short circuit current to a desired value 9. To determine the short circuit capability of series capacitors used in series compensation of long transmission lines 10. To determine the size of grounding transformers, resistances, or reactors
Short Circuit Calculations IIEE Presentation
Per Unit Calculations
Short Circuit Calculations IIEE Presentation
Threephase Systems (base voltage, kVL− L ) × 1000 ZB = base kVA3Φ 2
(base voltage, kVL− L ) ZB = base MVA3Φ
Short Circuit Calculations IIEE Presentation
2
Per Unit Quantities I pu
actual current = Base Current ( I B )
actual voltage (kV ) V pu = Base Voltage ( kVB ) Z pu
Short Circuit Calculations IIEE Presentation
actual impedance = Base impedance (Z B )
Changing the Base of Per Unit Quantities Z pu[ old ]
actual impedance, Z (Ω) = 2 (base kV[old ] ) ×1000 base kVA[ old ] Z pu[ old ] (base kV[ old ] ) ×1000 2
Z (Ω ) =
base kVA[ old ]
( base kV ) ×1000 = 2
Z B[ new]
[ new ]
base kVA[ new]
Z pu[ new] Short Circuit Calculations IIEE Presentation
Z (Ω ) = Z B[ new]
Changing the Base of Per Unit Quantities Z pu [ old ]
actual impedance, Z (Ω) = (base kV[old ] )2 × 1000 base kVA[ old ] Z pu[ old ] (base kV[ old ] ) ×1000 2
Z ( Ω) =
base kVA[ old ]
( base kV ) ×1000 = 2
Z B[ new]
[ new ]
base kVA[ new]
Z pu[ new] Short Circuit Calculations IIEE Presentation
Z ( Ω) = Z B[ new]
kVA Base for Motors kVA/hp 1.00
Induction < 100 hp
1.00
Synchronous 0.8 pf
0.95
Induction 100 < 999 hp
0.90
Induction > 1000 hp
0.80
Synchronous 1.0 pf
Short Circuit Calculations IIEE Presentation
hp rating
SYMMETRICAL COMPONENTS
Short Circuit Calculations IIEE Presentation
Va 2
Va1
Vc1
Vb 2
Vb1
Positive Sequence
Vc 2
Negative Sequence Va 0
Va 0 = Vb 0 = Vc 0
Va Vc 0 Vc1
Vc
Va 2
Va1
Vc 2 Vb0
Vb Vb1
Zero Sequence Short Circuit Calculations IIEE Presentation
Vb2
Unbalanced Phasors
Symmetrical Components of Unbalanced Threephase Phasor Va = Va 0 + Va1 + Va 2 Vb = Va 0 + a Va1 + aVa 2 2
Vc = Va 0 + aVa1 + a Va 2 2
Short Circuit Calculations IIEE Presentation
1 Va 0 = (Va + Vb + Vc ) 3 1 Va1 = (Va + aVb + a 2Vc ) 3 1 Va 2 = (Va + a 2Vb + aVc ) 3
Symmetrical Components of Unbalanced Threephase Phasor In matrix form:
Va 1 1 1 Va 0 2 V = 1 a a V b a1 Vc 1 a a 2 Va 2
Short Circuit Calculations IIEE Presentation
1 1 1 Va Va 0 V = 1 1 a a 2 V b a1 3 1 a 2 a Vc Va 2
Power System Short Circuit Calculations Sequence Networks
Short Circuit Calculations IIEE Presentation
Fault Point The fault point of a system is that point to which the unbalanced connection is attached to an otherwise balanced system.
Short Circuit Calculations IIEE Presentation
Definition of Sequence Networks Positivesequence Network Ea1 = Thevenin’s equivalent voltage as seen at the fault point Z1 = Thevenin’s equivalent impedance as seen from the fault point
Va1 = Ea1 − I a1Z1 Short Circuit Calculations IIEE Presentation
Ia1
+ Z1 Ea1
Va1 
Definition of Sequence Networks Negativesequence Network Z2 = Theveninâ€™s equivalent negativesequence impedance as seen at the fault point
Ia2
+
Va 2 = âˆ’ I a 2 Z 2
Short Circuit Calculations IIEE Presentation
Z2
Va2 
Definition of Sequence Networks Zerosequence Network Z0 = Theveninâ€™s equivalent zerosequence impedance as seen at the fault point
Va 0 = âˆ’ I a 0 Z 0
Ia0
+ Z0
Va0 
Short Circuit Calculations IIEE Presentation
Power System Short Circuit Calculations Sequence Network Models of Power System Components
Short Circuit Calculations IIEE Presentation
Synchronous Machines (Positive Sequence Network)
Z1 = jx' 'd
Short Circuit Calculations IIEE Presentation
Synchronous Machines (Negative Sequence Network) Ia2
x ' 'd + x ' 'q Z2 = j 2
+ Z2
Va2 
Where:
x”d = directaxis subtransient reactance x”q = quadratureaxis subtransient reactance Short Circuit Calculations IIEE Presentation
Synchronous Machines (Zero Sequence Network) SolidlyGrounded Neutral
jX0 n I0 ground
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Synchronous Machines (Zero Sequence Network) ImpedanceGrounded Neutral jX0 n
3Zg ground
Short Circuit Calculations IIEE Presentation
I0
Synchronous Machines (Zero Sequence Network) UngroundedWye or Delta Connected Generators jX0 n
I0 ground Short Circuit Calculations IIEE Presentation
TwoWinding Transformers (Positive Sequence Network) P
S
P
S ZPS
Standard Symbol
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Equivalent Positiveseq. Network
ThreeWinding Transformers (Positive Sequence Network) ZS Zp S P T Standard Symbol
Z ps = Z p + Z s Z pt = Z p + Z t Z st = Z s + Z t Short Circuit Calculations IIEE Presentation
S
P Zt Equivalent PositiveSequence Network
1 Z p = (Z ps + Z pt − Z st ) 2 1 Z s = (Z ps + Z st − Z pt ) 2 1 Z t = (Z pt + Z st − Z ps ) 2
T
Transformers (Negative Sequence Network) The negativesequence network of twowinding and threewinding transformers are modeled in the same way as the positivesequence network since the positivesequence and negativesequence impedances of transformers are equal.
Short Circuit Calculations IIEE Presentation
Simplified Derivation of Transformer ZeroSequence Circuit Modeling (Thanks to Engr. Antonio C. Coronel, Retired VP, Meralco, an d form er m em ber, Board of Electrical Engineering)
P
Q
Grounded wye
S1 = 1 and S3 = 0 or S2 = 1 or S4 = 0
Delta
S1 = 0 and S3 = 1 or S2 = 0 and S4 = 1
Ungrounded wye
S1 = 0 and S3 = 0 or S2 = 0 and S4 = 0
Short Circuit Calculations IIEE Presentation
Simplified Derivation of Transformer ZeroSequence Circuit Modeling Grounded wye â€“ Grounded wye S1 = 1 S2 = 1 S3 = 0 S4 = 0 Z
P
S1
S2 S3
Short Circuit Calculations IIEE Presentation
S4
Q
Simplified Derivation of Transformer ZeroSequence Circuit Modeling Grounded wye â€“ Ungrounded wye S1 = 1 S2 = 0 S3 = 0 S4 = 0
P
Short Circuit Calculations IIEE Presentation
Q
Simplified Derivation of Transformer ZeroSequence Circuit Modeling Grounded wye â€“ Delta S1 = 1 S2 = 0 S3 = 0 S4 = 1
P
Short Circuit Calculations IIEE Presentation
Q
Simplified Derivation of Transformer ZeroSequence Circuit Modeling Delta â€“ Delta S1 = 0 S2 = 0 S3 = 0 S4 = 0
P
Short Circuit Calculations IIEE Presentation
Q
Transformers (ZeroSequence Circuit Model) ZeroSequence Circuit Equivalent
Transformer Connection
P P
Q
Q ZPQ
P P
Short Circuit Calculations IIEE Presentation
Q
Q ZPQ
Transformers (ZeroSequence Circuit Model) ZeroSequence Circuit Equivalent
Transformer Connection
P P
Q
Q ZPQ
P P
Short Circuit Calculations IIEE Presentation
Q
Q ZPQ
Transformers (ZeroSequence Circuit Model) ZeroSequence Circuit Equivalent
Transformer Connection P P
Q
ZPQ
P P
IIEE Presentation
ZP
Q R
Short Circuit Calculations
Q
ZQ ZR
R
Q
Transformers (ZeroSequence Circuit Model) ZeroSequence Circuit Equivalent
Transformer Connection
ZP
P P
ZQ
Q
ZQ
Q
Q ZR R
R
P P
Q R
Short Circuit Calculations IIEE Presentation
ZP ZR R
Transmission Lines (Positive Sequence Network) Z1 ra
Short Circuit Calculations IIEE Presentation
jXL
Transmission Lines (Negative Sequence Network) The same model as the positivesequence network is used for transmission lines inasmuch as the positivesequence and negativesequence impedances of transmission lines are the same
Short Circuit Calculations IIEE Presentation
Transmission Lines (Zero Sequence Network) The zerosequence network model for a transmission line is the same as that of the positive and negativesequence networks. The sequence impedance of the model is of course the zerosequence impedance of the line. This is normally higher than the positive and negativesequence impedances because of the influence of the earthâ€™s resistivity and the ground wire/s. Short Circuit Calculations IIEE Presentation
Power System Short Circuit Calculations Classification of Power System Short Circuits
Short Circuit Calculations IIEE Presentation
Shunt Faults Single linetoground faults Double linetoground faults Linetoline faults Threephase faults
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Series Faults
Oneline open faults Twoline open faults
Short Circuit Calculations IIEE Presentation
Combination of Shunt and Series Faults Single linetoground and oneline open Double linetoground and oneline open faults Linetoline and oneline open faults Threephase and oneline open faults
Short Circuit Calculations IIEE Presentation
Combination of Shunt and Series Faults Single linetoground and twoline open faults Double linetoground and twoline open faults Linetoline and twoline open faults Threephase and twoline open faults
Short Circuit Calculations IIEE Presentation
Balanced Faults Symmetrical or ThreePhase Faults
Short Circuit Calculations IIEE Presentation
Derivation of Sequence Network Interconnections A Ia
B Ib
C Ic
System A Zf Va
Vb
Zf
System B
Zf
Vc
Vf
Boundary conditions: Ia + Ib + Ic = 0 VF = Va − I a Z f = Vb − I b Z f = Vc − I c Z f Short Circuit Calculations IIEE Presentation
Eq’n (1) Eq’n (2)
Zf
Zf
Ia1
Ia0
I a2 + Z1 Ea1
+ Z2 Va1
Va2 

Ea1 I a1 = Z1 + Z f I f = I a 0 + I a1 + I a 2 Ea1 I f = I a1 = Z1 + Z f Zf = 0, Short Circuit Calculations IIEE Presentation
Ea 1 I f = I a1 = Z1
+ Z0
Va0 
Unbalanced Faults Single LinetoGround Faults
Short Circuit Calculations IIEE Presentation
Derivation of Sequence Network Interconnections A Ia
B Ib Ic
System A Zf Va
Vb
Vc
Boundary conditions: Ib = Ic = 0 Va = I a Z f = 0 Short Circuit Calculations IIEE Presentation
C System B
Ia1
I a 0 = I a1 = I a 2
+ Z1 Ea1
Va1
If Z f = 0

Ia2
+ Z2
Ea1 = Z 0 + Z1 + Z 2 + 3Z f
3Zf
I a 0 = I a1 = I a 2
E a1 = Z 0 + Z1 + Z 2
I a 0 = I a1 = I a 2
Ea1 = Z 0 + 2 Z1
Z1 = Z 2
Va2 
Ia0
I f = I a = I a 0 + I a1 + I a 2 = 3I a1 = 3I a 0 + Z0
Va0
If Z f = 0 and Z1 = Z 2

3Ea1 I f = Ia = Z 0 + 2 Z1 Short Circuit Calculations IIEE Presentation
Unbalanced Faults LinetoLine Faults
Short Circuit Calculations IIEE Presentation
Derivation of Sequence Network Interconnections A Ia
B Ib Ic
System A Va
Vb
Vc
Zf
Boundary conditions: Ia = 0 Ib = −Ic Vb − I b Z f = Vc ; or Vb − Vc = I b Z f Short Circuit Calculations IIEE Presentation
C System B
Zf Ia1 Ia2
Ia0
+ Z1 Ea1
+ Va1
Z2
+ Z0
Va2 
Va0 

E a1 I a1 = Z1 + Z 2 + Z f If Z f = 0 and Z1 = Z 2 E a1 I a1 = 2 Z1 Short Circuit Calculations IIEE Presentation
The fault current I f = Ib = −I c = I a 0 + a 2 I a1 + aI a 2 I ao = 0; I a1 = − I a 2
(
)
I f = a 2 − a I a1 = − j 3I a1
thus, with Z1 = Z 2 Ea 1 I f = − j 3 2Z1 + Z f if Z f = 0
3 Ea1 3 Ea1 If = − j = Z 2Z 1 2 1 Ea1 I f [3φ ] = Z1 I f [ L −L ] Short Circuit Calculations IIEE Presentation
3 = I f [3φ ] 2
Unbalanced Faults Doubletoline Ground Fault
Short Circuit Calculations IIEE Presentation
Derivation of Sequence Network Interconnections A Ia
B Ib
C Ic
System A Zf Va
Vb
Zf
Vc
Vf
Zg
(Ib+Ic)
Boundary conditions: Ia = 0
Vb = I b Z f + (I b + I c )Z g
Eq’n BC2
Vc = I c Z f + (I b + I c )Z g
Eq’n BC3
Short Circuit Calculations IIEE Presentation
Eq’n BC1
System B
Zf
Zf
Z f+3Z g
Ia1 Ia2
Ia0
+ Z1
+ Z2
Ea1
Va1
Va2 

I a1 = Z1 + Z f + Short Circuit Calculations IIEE Presentation
Ea1 (Z2 + Z f )(Z 0 + Z f + 3Z g ) Z 2 + Z 0 + 2Z f + 3Z g
+ Z0
V a0 
Negativesequence Component: Z 0 + Z f + 3Z g = − I a1 Z + Z + 2 Z + 3 Z 0 f g 2
Ia 2
Zerosequence Component: Ia0
Z2 + Z f = − I a1 Z + Z + 2 Z + 3Z 0 f g 2
The fault current
I f = I b + I c = (I a 0 + a 2 I a1 + aI a 2 ) + (I a 0 + aI a1 + a 2 I a 2 ) I f = 2 I a 0 + (a 2 + a )I a1 + (a + a 2 )I a 2
I f = 2 I a 0 + (− 1)I a1 + (− 1)I a 2 = 2 I a 0 − (I a1 + I a 2 ) but I a 0 + I a1 + I a 2 = 0; or I a 0 = −(I a1 + I a 2 ) thus, I f = 3I a 0 Short Circuit Calculations IIEE Presentation
If Z f = Z g = 0 and Z1 = Z 2 Ea 1 ( Z1 + Z 0 )Ea1 I a1 = = 2 Z1Z 0 Z + 2 Z Z 1 1 0 Z1 + Z1 + Z 0
I a2
Z0 Ea1 = − = − I a1 Z1Z 0 Z1 + Z 0 Z1 + Z + Z 1 0 Z 0 Ea1 − 2 Z1 + 2 Z1Z 0
Short Circuit Calculations IIEE Presentation
Z 0 = Z1 + Z 0
If Z f = Z g = 0 and Z1 = Z 2 I a1 =
Ia2
Ea 1 (Z + Z 0 )Ea1 = 12 ZZ Z1 + 2 Z1Z 0 Z1 + 1 0 Z1 + Z 0
Z0 Ea1 = − I a1 =− Z1Z0 Z1 + Z0 Z1 + Z + Z 1 0 Z 0 Ea 1 − 2 Z1 + 2 Z1Z 0
Z 0 = Z1 + Z 0
Z1 Ea1 Z1 I a 0 = −I a1 = Z1 + Z 0 Z + Z1 Z0 Z1 + Z 0 1 Z +Z 1 0 ZE Ea1 = 2 1 a1 = Z1 + 2 Z1 Z 0 Z1 + 2 Z 0 I f = 3I a 0 = Short Circuit Calculations IIEE Presentation
3 Ea1 Z1 + 2 Z 0
Voltage Rise Phenomenon Singletoline Ground Fault
Short Circuit Calculations IIEE Presentation
Unfaulted Phase B Voltage During Single LinetoGround Faults Vb = Va0 + a 2Va1 + aVa 2 Ea 1 E a1 Ea1 2 Z 0 + a Ea1 − Z1 − a Z1 Vb = − 2 Z1 + Z 0 2 Z1 + Z 0 2 Z1 + Z 0 Z0 − 1 2 Z 0 − Z1 Z Z = Ea1 a 2 − 1 1 Vb = Ea1 a − 2 Z1 + Z 0 Z1 2 + Z 0 Z 1 Z0 − 1 Z Vb = Ea1 a 2 − 1 Z 0 2 + Z 1 neglecting resistances, R 1 and R 0 ; X0 − 1 X Vb = Ea1 a 2 − 1 X 0 Short Circuit Calculations 2 + X IIEE Presentation 1
Unfaulted Phase B Voltage During Single LinetoGround Faults Neglecting resistances R0 & R1 50 45 40
Phase B Voltage (p.u.)
35
30
25 20
15
10 5
0 10
6
3
2.8
2.6
2.4
2.2
2
1.8
1.6
1.4
1.2 X0/X1
Short Circuit Calculations IIEE Presentation
1
0
1
2
5
9
25
45
65
85
Fault MVA MVAF = I F × MVAbase where, for E
a1
= 1.0 p.u.;
for three − phase fault in p.u. :
I F ( 3φ )
1 = Z1
for singe line  to  ground fault in p.u. :
I F ( SLG ) Short Circuit Calculations IIEE Presentation
3 = Z 0 + 2Z1
Fault MVA Three − phase fault MVA : MVAF ( 3φ ) = I F (3φ ) ( p.u.) × MVAbase Z1 =
1
p.u.
I F (3φ )
Single line − to − ground fault MVA : MVAF ( SLG ) = I F ( SLG ) ( p.u.) × MVAbase 2 Z1 + Z 0 = Z0 = Short Circuit Calculations IIEE Presentation
3 I F ( SLG )
3 I F ( SLG )
− 2Z 1
p.u.
Assumptions Made to Simplify Fault Calculations 1. Prefault load currents are neglected. 2. Prefault voltages are assumed equal to 1.0 per unit. 3. Resistances are neglected (only for 115kV & up). 4. Mutual impedances, when not appreciable are neglected. 5. Offnominal transformer taps are equal to 1.0 per unit. 6. Positive and negativesequence impedances are equal. Short Circuit Calculations IIEE Presentation
Outline of Procedures for Short Circuit Calculations 1 Setup the network impedances expressed in per unit on a common MVA base in the form of a singleline diagram 2 Determine the single equivalent (Theveninâ€™s) impedance of each sequence network. 3 Determine the distribution factor giving the current in the individual branches for unit total sequence current.
Short Circuit Calculations IIEE Presentation
Outline of Procedures for Short Circuit Calculations 4 Interconnect the three sequence networks for the type of fault under considerations and calculate the sequence currents at the fault point. 5 Determine the sequence current distribution by the application of the distribution factors to the sequence currents at the fault point 6 Synthesize the phase currents from the sequence currents.
Short Circuit Calculations IIEE Presentation
Outline of Procedures for Short Circuit Calculations 7 Determine the sequence voltages throughout the networks from the sequence current distribution and branch impedances 8 Synthesize the phase voltages from the sequence voltage components 9 Convert the pre unit currents and voltages to actual physical units
Short Circuit Calculations IIEE Presentation
CIRCUIT BREAKING SIZING (Asymmetrical Rating Factors) Momentary Rating Multiplying Factor = 1.6
Interrupting Rating Multiplying Factor 8 cycles 5 cycles 3 cycles 1 Â˝ cycles Short Circuit Calculations IIEE Presentation
= 1.0 = 1.1 = 1.2 = 1.5
EXAMPLE PROBLEM In the power system shown, determine the momentary and interrupting ratings for primary and secondary circuit breakers of transformer T2.
Short Circuit Calculations IIEE Presentation
Solution: Equivalent 69kV system@ 100MVA :
T1 :
800 I F (3 φ ) = = 8 pu 100 1000 I F ( SLG ) = = 10 pu 100 1 x1 = = 0.125 pu 8 3 x0 = − 2 × 0.125 = 0.05 pu 10
100 x = 0 .08 = 0.2667 pu 30 G1 : 100 x" = 0.10 = 0.40 pu 25 100 x0 = 0.06 = 0.24 pu 25 T2, T3, T3 : 100 x = 0 .06 = 0.80 pu 75 M1, M2 : kVAB = 0.9 × 5000 = 4500 or MVAB = 4.5
Short Circuit Calculations IIEE Presentation
100 x = 0 .15 = 3 .3333 pu 4.5
Solution: Positive sequence network:
Short Circuit Calculations IIEE Presentation
Solution: Zero sequence network:
Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models
Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models
Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models
Short Circuit Calculations IIEE Presentation
Incorrect Sequence Network Models
Short Circuit Calculations IIEE Presentation
QUESTIONS?
Short Circuit Calculations IIEE Presentation
Short Circuit Calculations IIEE Presentation
Published on Feb 11, 2013
2013,techpres,shortcircuit,calculations