Short Circuit Calculations

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 FAULT-PROOF 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


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


Per Unit Calculations


Three-phase Systems (base voltage, kVL− L ) × 1000 ZB = base kVA3Φ 2

(base voltage, kVL− L ) ZB = base MVA3Φ


2

Per Unit Quantities I pu

actual current = Base Current ( I B )

actual voltage (kV ) V pu = Base Voltage ( kVB ) Z pu


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


hp rating

SYMMETRICAL COMPONENTS


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 Three-phase Phasor Va = Va 0 + Va1 + Va 2 Vb = Va 0 + a Va1 + aVa 2 2

Vc = Va 0 + aVa1 + a Va 2 2


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 Three-phase 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 


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


Fault Point The fault point of a system is that point to which the unbalanced connection is attached to an otherwise balanced system.


Definition of Sequence Networks Positive-sequence 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 Negative-sequence Network Z2 = Thevenin’s equivalent negative-sequence impedance as seen at the fault point

Ia2

+

Va 2 = − I a 2 Z 2


Z2

Va2 -

Definition of Sequence Networks Zero-sequence Network Z0 = Thevenin’s equivalent zero-sequence impedance as seen at the fault point

Va 0 = − I a 0 Z 0

Ia0

+ Z0

Va0 -


Power System Short Circuit Calculations Sequence Network Models of Power System Components


Synchronous Machines (Positive Sequence Network)

Z1 = jx' 'd


Synchronous Machines (Negative Sequence Network) Ia2

x ' 'd + x ' 'q Z2 = j 2

+ Z2

Va2 -

Where:

x”d = direct-axis sub-transient reactance x”q = quadrature-axis sub-transient reactance Short Circuit Calculations IIEE Presentation

Synchronous Machines (Zero Sequence Network) Solidly-Grounded Neutral

jX0 n I0 ground


Synchronous Machines (Zero Sequence Network) Impedance-Grounded Neutral jX0 n

3Zg ground


I0

Synchronous Machines (Zero Sequence Network) Ungrounded-Wye or Delta Connected Generators jX0 n

I0 ground Short Circuit Calculations IIEE Presentation

Two-Winding Transformers (Positive Sequence Network) P

S

P

S ZPS

Standard Symbol


Equivalent Positive-seq. Network

Three-Winding 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 Positive-Sequence 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 negative-sequence network of twowinding and three-winding transformers are modeled in the same way as the positive-sequence network since the positive-sequence and negative-sequence impedances of transformers are equal.


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


Simplified Derivation of Transformer Zero-Sequence Circuit Modeling Grounded wye – Grounded wye S1 = 1 S2 = 1 S3 = 0 S4 = 0 Z

P

S1

S2 S3


S4

Q

Simplified Derivation of Transformer Zero-Sequence Circuit Modeling Grounded wye – Ungrounded wye S1 = 1 S2 = 0 S3 = 0 S4 = 0

P


Q

Simplified Derivation of Transformer Zero-Sequence Circuit Modeling Grounded wye – Delta S1 = 1 S2 = 0 S3 = 0 S4 = 1

P


Q

Simplified Derivation of Transformer Zero-Sequence Circuit Modeling Delta – Delta S1 = 0 S2 = 0 S3 = 0 S4 = 0

P


Q

Transformers (Zero-Sequence Circuit Model) Zero-Sequence Circuit Equivalent

Transformer Connection

P P

Q

Q ZPQ

P P


Q

Q ZPQ



P P

Q

Q ZPQ

P P


Q

Q ZPQ


Transformer Connection P P

Q

ZPQ

P P

IIEE Presentation

ZP

Q R

Short Circuit Calculations

Q

ZQ ZR

R

Q



ZP

P P

ZQ

Q

ZQ

Q

Q ZR R

R

P P

Q R


ZP ZR R

Transmission Lines (Positive Sequence Network) Z1 ra


jXL

Transmission Lines (Negative Sequence Network) The same model as the positive-sequence network is used for transmission lines inasmuch as the positive-sequence and negative-sequence impedances of transmission lines are the same


Transmission Lines (Zero Sequence Network) The zero-sequence network model for a transmission line is the same as that of the positive- and negative-sequence networks. The sequence impedance of the model is of course the zero-sequence impedance of the line. This is normally higher than the positive- and negative-sequence impedances because of the influence of the earth’s resistivity and the ground wire/s. Short Circuit Calculations IIEE Presentation

Power System Short Circuit Calculations Classification of Power System Short Circuits


Shunt Faults Single line-to-ground faults Double line-to-ground faults Line-to-line faults Three-phase faults


Series Faults

One-line open faults Two-line open faults


Combination of Shunt and Series Faults Single line-to-ground and one-line open Double line-to-ground and one-line open faults Line-to-line and one-line open faults Three-phase and one-line open faults


Combination of Shunt and Series Faults Single line-to-ground and two-line open faults Double line-to-ground and two-line open faults Line-to-line and two-line open faults Three-phase and two-line open faults


Balanced Faults Symmetrical or Three-Phase Faults


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 Line-to-Ground Faults



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 Line-to-Line Faults



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 Double-to-line Ground Fault



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 BC-2

Vc = I c Z f + (I b + I c )Z g

Eq’n BC-3


Eq’n BC-1

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 -

Negative-sequence Component:  Z 0 + Z f + 3Z g   = − I a1   Z + Z + 2 Z + 3 Z 0 f g   2

Ia 2

Zero-sequence 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


   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 Single-to-line Ground Fault


Unfaulted Phase B Voltage During Single Line-to-Ground 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 Line-to-Ground 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


-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. Pre-fault load currents are neglected. 2. Pre-fault 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. Off-nominal transformer taps are equal to 1.0 per unit. 6. Positive- and negative-sequence 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 single-line diagram 2 Determine the single equivalent (Thevenin’s) impedance of each sequence network. 3 Determine the distribution factor giving the current in the individual branches for unit total sequence current.


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.


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


CIRCUIT BREAKING SIZING (Asymmetrical Rating Factors) Momentary Rating Multiplying Factor = 1.6

Interrupting Rating Multiplying Factor 8 cycles 5 cycles 3 cycles 1 ½ 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.


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


100  x = 0 .15 = 3 .3333 pu   4.5 

Solution: Positive sequence network:


Solution: Zero sequence network:


Improper Sequence Network Models






Incorrect Sequence Network Models


QUESTIONS?



Short Circuit Calculations

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