Mack Grady Fundamentals of Electric Power TAMU Relay Conf Tutorial 150331

TOPICS
1. Power Definitions and Equations. Why?
As with any other technology, the underlying physics must be understood so that equations and models
can be developed that simulate actual behavior.
2. Three‐Phase Power. Why?
Losses are minimized because there are no ground or neutral return currents. Also, three‐phase machines
are smaller and much more efficient than single‐phase machines of the same power rating.
3. Transformer Models. Why?
To move power over long distances and minimize I²R losses, transmission line voltages must be boosted to
much higher levels than voltages produced by generators or required by end‐users.
4. Per Unit System. Why ?
To make modeling and simulation of circuits having transformers much easier. A by‐product is that per‐
unitized equipment parameters such as ratings and impedances fall into narrow and predictable ranges.
5. Symmetrical Components. Why?
Greatly simplifies analysis of normal power system operation, as well as abnormal events such as
unbalanced faults.
6. Transmission Line Models. Why?
Necessary for designing and simulating power systems for loadflow, short‐circuit, and stability purposes.
7. Fault (Short‐Circuit) Calculations and Voltage Sags. Why?
So that a power system can be properly protected by detecting and isolating faults within 0.1 second. 1
second is a very long time for a fault to exist ‐ long enough to cause grid blackouts.
Prof. Mack Grady, TAMU Relay Conference
1
Tutorial, Topic 1, March 31, 2015
1. Power Definitions and Equations, cont.
As with any other technology, the underlying physics must be understood so that equations and models can
be developed that simulate actual behavior.
Instantaneous power p(t) flowing into the box i (t )
v(t )  V sin(ot   ), +
Circuit in a box,
i (t )  I sin(ot   ) v (t ) two wires
p (t )  v(t )  i (t ) −
i (t )
p (t )  v(t )  i (t )  V sin(ot   )  I sin(ot   )
zero average
 cos(   )  cos(2ot     ) 
p(t )  VI   peak
 2 
1 t o T VI V I
Pavg 
T 
to
p (t )dt 
2
cos(   ) 
2 2
cos(   )
Average power Pavg flowing into the box
rms
Pavg  Vrms I rms cos(   )
Power factor
2
1. Power Definitions and Equations.
As with any other technology, the underlying physics must be understood so that equations and
models can be developed that simulate actual behavior.
i R (t )  v R (t )  ,
v (t ) ,
iR (t )  R ∴ in phase with
R
 v L (t )  ,
i L (t ) di (t )
vL (t )  L ∴
,
dt lags by 90°
iC (t )  vC (t )  ,
dv(t )
iC (t )  C ,
dt ∴ leads by 90°
3
Thanks to Charles Steinmetz, Steady‐State AC problems are greatly simplified with
phasor analysis because differential equations are replaced by complex numbers
Time Domain Frequency Domain
~
v (t ) VR
Resistor i R (t )  R ZR  ~  R
R IR
voltage leads current
di ( t ) ~
VL
Inductor v L (t )  L Z L  ~  j L
dt IL
current leads voltage
dv ( t ) ~
iC ( t )  C VC 1
Capacitor ZC  ~ 
dt IC j C
4
Voltage and Current Phasors for R’s, L’s, C’s
~ Voltage and
VR ~ ~
Resistor Z R  ~  R , V R  RI R Current in phase Q = 0
IR
Voltage leads
~ Current by 90°
VL ~ ~
Inductor Z L  ~  j  L, V L  j  LI L Q > 0
IL
~ ~ Current leads
VC 1 ~ IC Voltage by 90°
Capacitor ZC  ~  , VC  Q < 0
IC j C j C
5
Converting Time Domain Waveforms to Phasor Domain
Using a cosine reference,
Voltage cosine has peak = 100V, phase angle = ‐90º
Current cosine has peak = 50A, phase angle = ‐135º
~ 100 ~ 50
Phasors V    90 V , I    135 A
2 2
100
V I
80 Pavg   cos(   )
2 2
60
100 50
40 Pavg   cos 90  (135) 
20 2 2
Voltage
0
Current
0 30 60 90 120 150 180 210 240 270 300 330 360
-20
100 50
-40 Pavg   cos45
-60 2 2
-80 Pavg  1767W
-100
6
Circuit analysis using the Nodal Method. Write KCL equations at major nodes 1 and 2,
and solve for phasor voltages V1 and V2 .
~ ~
V1 V2
Reference
10020 1 
  
1 1 1 1  10020  
j4 2

 j4  3  2 
2  V~   j4   100  20 1
1
1 
   ~1     ~  1 2  j 2 
 2  10020 
 1 1 1 V V1 
 1 D
 2 2  j 2   1 
 1 10020 
 
 1 1 1 1 1   1  1 2 j4 
D        1      
 j4 3 2   2  j 2   2   2  1 1 1 10020 
~ 2  j2 1 
V2  
D
7
Active power Pavg and reactive power Q form a power triangle
V I V I
Pavg  cos(   ), Q  sin(   ),
2 2 2 2
~ ~*

S  P  jQ  V  I  V  I   VI   
*
~
V  V Complex power
Projection of S
(   ) S Q on the
imaginary axis
~ (   )
I  I 
P
Projection of S on the real axis
cos(   ) is the power factor
8
Resistor
~ *

~ V  V 2 V 2
S  P  jQ  V     * 
Z  Z R Thus
V2
Alternatively, P  I 2 R, Q  0
R
~
 
~*
S  P  jQ  I Z  I  I 2 Z  I 2 R
Inductor ~ *

~ V  V 2
S  P  jQ  V     * 
V2
 jL
 j
V2
L
Z  Z Thus
V2
Alternatively, P  0, Q   LI 2
~
 
~*
S  P  jQ  I Z  I  I 2 Z  jLI 2
L
Inductor consumes reactive power
Capacitor ~ *

~ V  V 2
S  P  jQ  V     * 
V2
1
  jCV 2
Z  Z
Alternatively,  j C Thus
I2
P  0, Q  CV 2 
~
 
~*
S  P  jQ  I Z  I  I 2
1
j C
  jLI 2
C
Capacitor produces reactive power
Always use rms values of voltage and current in the above equations
Tutorial, Topic 1, March 31, 2015 9
Question: Why is the sum of power out IB
of a node = 0? IA
Answer: KCL and conservation of power
Question: What about reactive power Q? IC
Answer: It depends.
Question: Can you be a bit more specific? ~ ~ ~
Answer: Unlike P, there is no physical for I A  I B  IC  0
Q to be conserved.
When voltage and current are not
~ ~
 ~ ~
V I A  I B  IC  0 
 
sinusoidal, then cross products of voltage ~~ ~ ~ *
and current exist and Q is not conserved. V I A  I B  IC  0
But power systems are mostly sinusoidal, PA  jQ A  PB  jQB  PC  jQC  0
so as shown on the right with phasors,
both P and Q are conserved. PA  PB  PC  0
Q A  QB  QC  0
10
2. Three‐Phase Power.
i A (t )
Instantaneous power p(t) flowing into the box,  v A (t )
p ABC (t )  v A (t )  i A (t )  vB (t )  iB (t )  vC (t )  iC (t ) iB (t )
 vB (t ) Circuit in a box,
iC (t ) four‐wires
For a balanced system,
 vC (t )
p ABC (t )  VA sin(t   )  I A sin(t   ) iN (t )
v N (t )  0
 V A sin(t    120)  I A sin(t    120)
 V A sin(t    120)  I A sin(t    120).
Combining terms,
p ABC (t )  VA I A sin(t   ) sin(t   )
 sin(t    120) sin(t    120)
 sin(t    120) sin (t    120).
cos(   )  cos(   )
sin( ) sin(  ) 
Trig. identity yields
2
VAI A
pABC(t)  cos(  )  cos(2t    )  cos(  )  cos(2t     240)  cos(  )  cos(2t     240) ,
2
1
2. Three‐Phase Power, cont.
VA I A  cos(2t     )  cos(2t      240)  cos(2t      240) 

p ABC (t )  3  cos(   )  .
2 2 3
Letting x = (2ωt+δ+Θ) and expanding the time varying term,
cos( x)  cos( x  240)  cos( x  240)  cos( x)  cos( x  120)  cos( x  120),
cos( x)  cos( x  120)  cos( x  120)

then expanding yields
cancel each other
cos( x)  cos( x) cos(120)  sin( x) sin(120)  cos( x) cos(120)  sin( x) sin(120),
cos( x)  cos( x) cos(120)  cos( x) cos(120),
 1 1
cos( x)1  cos(120)  cos(120)  cos( x) 1     0,
 2 2
Instantaneous three‐phase
VA I A
p ABC (t )  3 cos(   )  3VArms I Arms cos(   ) power is constant – thus,
2 2 smooth running machines!
2
Vmag = 1 Analogous to a piston engine with
Vang = 0
Imag = 0.50 50 Strong 120 Hz torque an infinite number of cylinders
Iang = -30 150
Phase A Phase A Phase A P Q Phase B Phase B Phase B Phase C Phase C Phase C A+B+C Q
wt v(t) I(t) p(t) 0.216506 0.125 v(t) I(t) p(t) v(t) I(t) p(t) p(t) 0.375
0 1 0.433013 0.433013 0.216506 0.125 -0.5 -0.43301 0.216506 -0.5 3.06E-17 -1.5E-17 0.649519 0.375
2
Instantaneous Power in Single-Phase
0.999391 0.441474 0.441205 0.216506
Circuit
0.125 -0.46947 -0.42402 0.199067 Instantaneous Power in0.009247
-0.52992 -0.01745 Three-Phase
0.649519Circuit
0.375
1.5 4 0.997564 0.449397 0.448302 0.216506 0.125 -0.43837 -0.41452 1.5 0.181713 -0.55919 -0.03488 0.019504 0.649519 0.375
6 0.994522 0.456773 0.45427 0.216506 0.125 -0.40674 -0.40451 0.164528 -0.58779 -0.05226 0.03072 0.649519 0.375
8 0.990268 0.463592 0.45908 0.216506 0.125 -0.37461 -0.39401 0.147597 -0.61566 -0.06959 0.042842 0.649519 0.375 va
10 0.984808 0.469846 0.462708 0.216506 0.125 -0.34202 -0.38302 0.131001 -0.64279 -0.08682 0.055809 0.649519 0.375 ia
12 0.978148 0.475528 0.465137 0.216506 va
0.125 -0.30902 -0.37157 0.114822 -0.66913 -0.10396 0.06956 0.649519 0.375 vb
14 0.970296 0.480631 0.466354 0.216506 0.125 -0.27564 -0.35967 0.099138 -0.69466 -0.12096 0.084027 0.649519 0.375
ia ib
0 16 0.961262 0.485148 0.466354 0.216506 0.125 -0.24192 -0.34733 0.084027
0 -0.71934 -0.13782 0.099138 0.649519 0.375
18 0.951057 0.489074 0.465137 0.216506 0.125 -0.20791 -0.33457
pa 0.06956 -0.74314 -0.15451 0.114822 0.649519 0.375 vc
20 0.939693 0.492404 0.462708 0.216506 0.125 -0.17365 -0.32139 0.055809 -0.76604 -0.17101 0.131001 0.649519 0.375 ic
Q
22 0.927184 0.495134 0.45908 0.216506 0.125 -0.13917 -0.30783 0.042842 -0.78801 -0.1873 0.147597 0.649519 0.375 pa+pb+pc
24 0.913545 0.497261 0.45427 0.216506 0.125 -0.10453 -0.29389 0.03072 -0.80902 -0.20337 0.164528 0.649519 0.375
Q
26 0.898794 0.498782 0.448302 0.216506 0.125 -0.06976 -0.2796 0.019504 -0.82904 -0.21919 0.181713 0.649519 0.375
28 0.882948 0.499695 0.441205 0.216506 0.125 -0.0349 -0.26496 0.009247 -0.84805 -0.23474 0.199067 0.649519 0.375
-1.5 30 0.866025 0.5 0.433013 0.216506 0.125 6.13E-17 -0.25-1.5-1.5E-17 -0.86603 -0.25 0.216506 0.649519 0.375
032 90 180 270 360
0.848048 0.499695 0.423766 0.216506 450 540 630 720 0 90 180 270 360
0.125 0.034899 -0.23474 -0.00819 -0.88295 -0.26496 0.233945 0.649519450 540 630 720
0.375
34 0.829038 0.498782 0.413509 0.216506 0.125 0.069756 -0.21919 -0.01529 -0.89879 -0.2796 0.2513 0.649519 0.375
3
Instantaneous Power to a Small Fan (medium power level)
200
150
100
Watts
50
-50
0.0000 0.0167 0.0333 0.0500
Seconds
Instantaneous Power to a Compact Fluorescent Lamp

200
150
100
Watts
50
-50
0.0000 0.0167 0.0333 0.0500
Seconds
4
Imaginary Balanced Three‐Phase System
Vcn Vab = Van – Vbn
Vca = Vcn – Van
30°
Real
Van
120°
Phasors rotate counter‐clockwise. Magnitude
of line‐to‐line voltage is √ times that of line‐
Vbn to‐neutral voltage.
Vbc =
Vbn – Vcn
5
Imaginary
Balanced Three‐Phase System
Vcn Vab = Van – Vbn
Vca = Vcn – Van
Ic
Ica
30°
Iab
Real
Ia Van
Line currents Ia, Ib, and Ic
Ib Ibc Delta currents Iab, Ibc, and Ica
I
c c
Ic
Vbn
Ica
Because line‐to‐line voltage magnitudes are Ibc
√ larger than line‐to‐neutral voltages,
conservation of power requires that
magnitudes of delta currents Iab, Ica, and Ibc Iab
Ib b
are smaller than Ia, Ib, Ic. a
√ Vbc =
– Vab +
Vbn – Vcn
Ia
6
c
c Ic
Ic
Z
3Z 3Z
n
Z Z
b a
b a
Ib Ib – Vab +
3Z
– Vab +
Ia Ia
Balanced three‐phase systems, no matter if they are delta connected, wye connected, or a
mix of wye and delta, are easier to solve if you follow these steps:
• Convert the entire circuit to an equivalent wye with a grounded neutral.
• Draw the one‐line diagram for phase a, recognizing that phase a has one third of the P
and Q.
• Solve the one‐line diagram for line‐to‐neutral voltages and line currents.
• If needed, compute line‐to‐neutral voltages and line currents for phases b and c using
the ±120° relationships.
• If needed, compute line‐to‐line voltages and delta currents using the and ±30°
relationships.
7
3. Transformer Models.
Turns ratio 7200:240

(30 : 1)
(approx. same amount of copper in each winding)
Rs jXs
Ideal
Rm jXm Transformer
7200:240V
7200V 240V
1
Tutorial Topic 3, March 31, 2015
3. Transformer Models, cont.
Open Circuit Test

Ioc
Rs jXs
Ideal +
Rm jXm Transformer Voc
7200:240V
-
7200V 240V
Φ Open circuit test: Open circuit

the 7200V-side, and apply
240V to the 240V-side. The
winding currents are small, so
the series terms are negligible.
~
V
Turns ratio 7200:240 Rm || jX m  ~oc
I oc
2
Short Circuit Test
Isc
Rs jXs
+ Ideal
Vsc Rm jXm Transformer
7200:240V
-
Short circuit test: Short circuit 7200V 240V
the 240V-side, and raise the
7200V-side voltage to a few
percent of 7200, until rated Φ
current flows. There is almost
no core flux so the
magnetizing terms are
negligible.
~
V
Rs  jX s  ~sc
I sc Turns ratio 7200:240
3
X / R Ratios for Three-Phase Transformers

• 345kV to 138kV, X/R = 10
• Substation transformers (e.g., 138kV to 25kV or 12.5kV, X/R = 2, X = 12%
• 25kV or 12.5kV to 480V, X/R = 1, X = 5%
• 480V class, X/R = 0.1, X = 1.5% to 4.5%
Rs jXs
Ideal
Rm jXm Transformer
4
1. Given the standard percentage values below for a 125kVA transformer, determine the R’s and
X’s in the diagram, in Ω.
2. If the R’s and X’s are moved to the 240V side, compute the new Ω values.
3. If standard open circuit and short circuit tests are performed on this transformer, what will be
the P’s and Q’s (Watts and VArs) measured in those tests?
Load
Single Phase Transformer. loss Xs
Percent values on
transformer base.
No-load
Magnetizing
loss
Winding 1 current
kV = 7.2, kVA = 125
Rs jXs
Winding 2 Ideal
kV = 0.24, kVA = 125
Rm jXm Transformer
%Imag = 0.5 7200:240V
%Load loss = 0.9
%No-load loss = 0.2 7200V 240V
%Xs = 2.2
5
EPRI Study, Distribution Feeder Loss Example
Annual energy loss = 2.40%
•
Largest component is transformer no-load
• loss (45% of the 2.40%)
Secondary Lines
21% Transformer No-
Load
Annual Feeder
Loss Components 45%
Primary Lines
26%
Transformer Load
8%
Modern Distribution Transformer:

• Load loss at rated load (I2R in conductors) = 0.75% of rated transformer kW.
• No load loss at rated voltage (magnetizing, core steel) = 0.2% of rated transformer kW.
• Magnetizing current = 0.5% of rated transformer amperes
6
A three-phase transformer can be three separate single-phase

transformers, or one large transformer with three sets of windings
Wye-Equivalent One-Line Model
Rs jXs
N1:N2 A Ideal
Rm jXm Transformer
N1 : N2
N1:N2
• Reflect to side 2 using individual transformer

turns ratio N1:N2
N1:N2 Standard 345/138kV autotransformers,GY-GY ,

Y-Y have a tertiary 12.5kV ∆ winding to permit
circulating 3 rd harmonic current
Prof. Mack Grady, TAMU Relay Conference 7
For Modeling a Delta-Delta Connection, Convert the Transformer to

Equivalent Wye-Wye
Rs jXs
A 3 Ideal
3
N1:N2 Rm jXm Transformer
3 3 N1 N2
:
N 3 3
N1:N2 • Convert side 1 impedances from delta to

equivalent wye
• Then reflect to side 2 using individual
transformer turns ratio N1:N2
N1:N2
∆-∆
8
For Modeling a Delta-Wye Connection, Convert the Transformer to

Equivalent Wye-Wye

Rs jXs
A 3 3 Ideal
N1:N2 Rm jXm Transformer
3 3 N1
: N2
N 3
N1:N2
• Convert side 1 impedances from delta to wye
• Then reflect to side 2 using three-phase line-to-line
turns ratio N 1 : 3 N 2
• Has 30° degree phase shift due to line-to-neutral to
N1:N2
line-to-line relationship. ANSI standard requires the
transformer to be labeled such that high-voltage
∆-Y side leads the low-voltage side by 30° .
9
For Modeling a Wye-Delta Connection, Convert the Transformer to

Equivalent Wye-Wye
A Rs jXs Ideal
N1:N2
Transformer
Rm jXm
N2
N1 :
N 3
N1:N2 • Reflect to side 2 using three-phase bank line-to-line

turns ratio 3 N 1 : N 2
• Has 30° degree phase shift due to line-to-neutral to
line-to-line relationship. ANSI standard requires the
transformer to be labeled such that high-voltage side
leads the low-voltage side by 30°
N1:N2
Y-∆
Thus, for all configurations, equivalent wye-wye transformer ohms can be reflected from
one side to the other using the three-phase bank line-to-line turns ratio
10
4. Per Unit System. Two equations
describe the ideal
Physical R’s, L’s. KCL, KVL. transformer
Rs jXs
Ideal
Rm jXm Transformer
From conservation of power,

I1 I2 v2 (t )  i2 (t )  v1 (t )  i1 (t )
+ +
V1 V2 v1 (t )
i2 (t )   i1 (t )
- - v2 (t )
Turns ratio N1 : N2 Merge Faraday’s Law into conservation of power,
N1
Faraday’s Law i2 (t )  i1 (t )
N2 N2
v2 (t )  v1 (t )  , N
v1 (t ) 2
N1 v2 (t ) N1 v1 (t )  N 2 
2
   
Conservation of Power i2 (t ) i (t ) N1 i1 (t )  N1 
v1 (t )  i1 (t )  v2 (t )  i2 (t )
1
N2
1
4. Per Unit System, cont.
2 Moving the impedances from low-

v2 (t ) v1 (t )  N 2 
   side to high-side AND multiplying
i2 (t ) i1 (t )  N1  them by the square of turns ratio
2
 N2 
 
Rs jXs  N1 
Ideal
Transformer jXm satisfies Faraday’s Law and
Rm
conservation of power, and provides
an exact equivalent circuit for the
transformer.
Turns ratio N1 : N2
• Developers of the per unit system recognized that an N1:N2 ideal transformer could be
replaced with a 1:1 ideal transformer if “base voltages” were used in circuit simulations,
where base voltages vary according to N1:N2 ratio.
• To achieve conservation of power and have the same base power everywhere in a
network, base current must vary according to the inverse of N1:N2 ratio.
• Thus, base voltage divided by base current varies according to (N2:N1)2, which means that
transformer impedances “in per unit” are identical on both sides of the transformer, and
the 1:1 ideal transformers can be excluded from circuit simulations.
2
2 Moving the impedances from low-

v2 (t ) v1 (t )  N 2 
   side to high-side AND multiplying
i2 (t ) i1 (t )  N1  them by the square of turns ratio
2
 N2 
 
Rs jXs  N1 
Ideal
Transformer jXm satisfies Faraday’s Law and
Rm
conservation of power, and provides
an exact equivalent circuit for the
transformer.
Turns ratio N1 : N2
• Developers of the per unit system recognized that an N1:N2 ideal transformer could be
replaced with a 1:1 ideal transformer if “base voltages” were used in circuit simulations,
where base voltages vary according to N1:N2 ratios.
• To achieve conservation of power and have the same base power everywhere in a
network, base current must vary according to the inverse of N1:N2.
• Thus, base voltage divided by base current varies according to (N2:N1)2, which means
that transformer impedances “in per unit” are identical on both sides of the
transformer, and the 1:1 ideal transformers are not needed in circuit simulations.
3
Single‐Phase Procedure
• Select an important bus in a system. Chose Vbase and Sbase values at that bus that fit
the conditions. For example, 7200 Vrms, 240 Vrms, etc. And, 25 kVA, 100 kVA, etc.
• Calculate Ibase(rms) = Sbase / Vbase. Zbase = Vbase / Ibase = (Vbase)2 / Sbase.
• If there are additional transformers in the circuit, adjust Vbase, Ibase, and Zbase for that
voltage level, so that Vbase varies with turns ratio, Ibase varies inversely with turns
ratio, and Zbase becomes the new Vbase/Ibase
• Replace transformers with their per unit internal impedances, expressed on the bases
you have chosen.
• Convert all impedances to per unit.
• Convert all P and Q to per unit using Sbase.
• Solve the circuit using per unit values.
• Once solved, convert per unit answers to actual volts, amps, voltamps by multiplying by
the appropriate base value.
4
Single‐Phase Procedure. Converting Per Unit Impedances from Equipment Base
to Your Chosen Base at their Bus
• Transformer impedances are given as per unit on the transformer base. Base voltage is
rated rms voltage, base voltamps is rated voltamps.
• Zpu = Zactual / Zbase, so Zactual = Zpu • Zbase
• Zactual = Zpu_old • Zbase_old. Also, the same Zactual = Zpu_new • Zbase_new
• Thus, Zpu_new = Zpu_old • Zbase_old / Zbase_new
• Continuing,
 2
V base, old   S base, new 
Z pu, new  Z pu, old   2 
S
 base, old  V base, new 
2
S  V 
Z  base, new   base, old 
pu , new  Z pu , old 
 S base, old  V base, new 
 
5
Extension to Three‐Phase
• The key to three‐phase per unit analysis is to recognize that transformer impedances
reflect from one side to the other in proportion to squared line‐to‐line voltage.
• In your mind, convert the system to a three‐phase wye‐equivalent with line‐to‐neutral
voltage. Per unit current is the current each wire, not the sum of all three. One wire
carries one‐third of the voltamps.
• Base voltage is line‐to‐line. Even so, it is helpful to remember its relationship to line‐to‐
neutral.
• Base voltamperes is the sum of three phases.
Vbase , LL
S base ,3  3S base ,1  3Vbase , LN I base  3 I base  3 Vbase , LL I base
3
2
Vbase , LL 
Z base ,3 
Vbase , LN 
2


 3 



Vbase , LL 
2
S base ,1 S base ,3 S base ,3

3
6
Converting Impedance from One Base to Another
2
Vbase , LL 
Z base ,3 
V
base , LN 
2 

3 
 V  2
 base , LL
S base ,1 S base ,3 S base ,3
3
Z actual  Z PU ,old  Z base ,old  Z PU ,new  Z base ,new
Z  V 2 baseLL ,3 ,old   S base ,3 ,new 

Z PU ,new  Z PU ,old  base ,old   Z PU ,old   2 
 Z base ,new   S base ,3 ,old  V baseLL ,3 ,new 
2
V   S base ,3 ,new 
Z PU ,new  Z PU ,old  baseLL ,3 ,old   
VbaseLL ,3 ,new   S base ,3 ,old 
7
Zero‐Sequence (i.e., Common Mode) Transformer Models
Grounded Wye - Grounded Wye R + jX
Grounded Wye - Delta R + jX
Grounded Wye - Ungrounded Wye

R + jX
Ungrounded Wye - Delta R + jX
Delta - Delta R + jX
8
Example Problem:
Information for a small power system is shown below. Per unit values are given on the equipment
bases. Using a 138kV, 100MVA base in the transmission lines, draw the positive and negative
sequence per unit diagrams. Assume that no current is flowing in the network, so that all
generator and motor voltages are 1.0pu (positive sequence) in your final diagram.
#1 #4 #5 #6 #3
Trans1 TLine1 TLine2 Trans3
Gen1 Trans1 Line1 Line2 Trans3 Motor

X” = 0.15 X = 0.16 R+ = 10Ω R+ = 10Ω X = 0.10 X” = 0.10
40MVA 60MVA X+ = 60Ω X+ = 60Ω 40MVA 25MVA
20kV GY, 138kV GY/ 138kV GY/ 13.2kV
j0.1 ohm 18kV Delta Trans2 13.8kV Delta Delta
grounding X=0.14 Trans2
reactor 50MVA
For transmission lines, assume that
138kV GY/ #2
R0 = 5 • R+
20kV Delta
X0 = 3 • X+
For generators and transformers,

Gen2, X” = 0.15, 35MVA, let Z0 = Z+
22kVGY, j0.1 ohm grounding
reactor Negative seq. impedances same as
positive seq. impedances.
9
5. Symmetrical Components.
Fortescue's Theorem: An unbalanced set of N related phasors can be resolved into N

systems of phasors called the symmetrical components of the original phasors. For a
three‐phase system (i.e. N = 3), the three sets are:
• Positive Sequence (indicated by “+” or “1”) ‐ three phasors, equal in magnitude,
120o apart, with the same sequence (a‐b‐c) as the original phasors.
• Nega ve Sequence (“−” or “2”) ‐ three phasors, equal in magnitude, 120o apart,

with the opposite sequence (a‐c‐b) of the original phasors.
• Zero Sequence (“0”) ‐ three identical phasors (i.e. equal in magnitude, with no

relative phase displacement).
The original set of phasors Symmetrical Components and their relationship to one another
~ ~ ~ ~ ~ ~ ~
Va  Va 0  Va1  Va 2 Va 0  Vb0  Vc0
~ ~ ~ ~ ~ ~ ~ ~
Vb  Vb0  Vb1  Vb 2 Vb1  Va1  1  120 o Vc1  Va1  1  120 o
~ ~ ~ ~ ~ ~ ~ ~
Vc  Vc0  Vc1  Vc 2 Vb 2  Va 2  1  120 o Vc 2  Va 2  1  120 o
5. Symmetrical Components, cont.
The symmetrical components of all a‐b‐c voltages are usually written in terms of the symmetrical
,
components of phase a by defining complex number a as
a  1  120 o
a 2  1  240 o  1  120 o
a 3  1  360 o  10 o
~ ~ ~
V a ,Vb ,Vc
Substituting into the previous equations for yields
~ ~ ~ ~ ~ ~
Va  Va 0  Va1  Va 2 Va  1 1 1  Va 0 
~   ~ 
a  Va1 
~ ~ ~ ~
Vb  Va 0  a 2Va1  aVa 2 Vb   1 a
2
~ ~ ~ ~ V~c  1 a ~
a 2  Va 2 
Vc  Va 0  aVa1  a 2Va 2  
which in simplified form is
1 1 1 1 1 1
~ ~ ~ ~ 1 1 
Vabc  T  V012 , V012  T 1  Vabc , T  1 a 2 
a  , T  1 a a 2 
 3
1 a a 2  1 a 2 a 
Why do all this? Answer: to simplify most situations.
~ ~
V012  T 1  Vabc ,
~ ~
Va 0  1 1 1  Va 
~  1 ~ 
Va1   3 1 a a 2  Vb 
V~a 2  1 a
2 ~
a 2  Vc 
 
For balanced set, have only positive‐sequence
~ ~
Va 0  1 1 1   Va  ~ 1 1 1  1  ~  1 a2  a  ~  1 a2  a  ~ 0  0 
~  1   ~ V 1 a V  3 Va  3 Va 3  V~ 1
Va1   3 1 a a 2  a 2Va   a  a 2  a 2   a 1  a  a   3
3
1  a  a   3
3
  a 
V~a 2  ~ 3 3
  1 a 2 a   aVa  1 a 2 a   a  1  a 4  a 2 
 
 1 a  a2 
  0 0
Swap phases b and c (negative rotation), have only negative‐sequence
~ ~
Va 0  1 1 1   Va  ~ 1 1 1  1  ~  1 a  a2  ~ 1  a 2  a  ~ 0  0 
~  1  ~  V V  4 Va   Va 0  V~ 0
Va1   3 1 a a 2   aVa   a 1 a
 a 2   a   a 1  a  a   3
2
1  a  a   3
2
  a 
V~a 2  ~ 3 3
  1 a 2 a  a 2Va  1 a 2 a  a 2  1  a 3  a 3 
 
 111 
  3 1
Common mode (a, b, c identical), have only zero‐sequence
~ ~
Va 0  1 1 1  Va  ~ 1 1 1  1 ~  3  ~  3 1
~  1  ~ V V 1  a  a 2   Va 0  V~ 0
Va1   3 1 a a 2  Va   a 1 a
 a 2  1  a   3   a 
V~a 2  ~ 3 3
  1 a 2 a  Va  1 a 2 a  1 1  a 2  a  0 0
Zero‐Sequence (i.e., Common Mode) Transformer Models, again
Zero‐sequence current can flow ONLY when there is a ground path
Grounded Wye - Grounded Wye R + jX
Grounded Wye - Ungrounded Wye R + jX
Delta - Delta R + jX
Three‐phase circuits can be solved in full a‐b‐c, or in using 0‐1‐2 sequence components.
For a balanced circuit, solve it in positive‐sequence because negative‐ and zero‐
sequences are zero, so that the positive‐sequence solution IS the phase A solution.
Hence, the “one‐line” diagram.
Consider the following a‐b‐c equations for a symmetric network. “Symmetric” means
that self‐impedances S for each phase are equal, and mutual‐impedances M between
phases are equal.
~ ~
Va   S M M I a 
~ ~ ~   ~ 
Vabc  Z abc  I abc
Vb    M S M   I b 
V~c   M M
~
S   I c 
 
Converting to 0‐1‐2
~ ~
TV012  Z abc  TI 012
~ ~
T 1TV012  T 1Z abcTI 012
~ ~
V012  Z 012 I 012 , where Z 012  T 1Z abcT
Z 012  T 1Z abcT

1 1 1  S M M  1 1 1
1
 1 a a 2   M S M  1 a 2 a 
3
1 a 2 a   M M S  1 a a 2 
S  2M S  2M S  2M  1 1 1
1
  S M a ( S  M ) a ( S  M ) 1 a 2
2
a 
3
 S  M a 2 ( S  M ) a ( S  M )  1 a a 2 
3( S  2M ) 0 0 
1 
  0 3( S  M ) 0 
3
 0 0 3( S  M )
This is huge! If network impedances
( S  2 M ) 0 0 
 are balanced, then a coupled a‐b‐c
 0 (S  M ) 0 
network can be solved using three
 0 0 ( S  M ) simple uncoupled 0‐1‐2 networks.
6. Transmission Line Models
Conductor with radius r, modeled electrically
Wire over Earth
as a line charge ql at the center
capacitance
b Electric Field
B ql
h E aˆ r Volts per meter
a 2 o r
A
Surface of Earth
bi ai
h
Image conductor, at an equal distance below

the Earth, and with negative line charge -ql
r b r  bi
ql   b   bi  ql  b  ai 
Vab   E   aˆ r   E i  aˆ r  2 o ln a   ln ai   2 o ln a  bi 
r a r  ai
ql  2h 
Vrg  ln  for h >> r.
2 o  r 
2π • 8.854 pF per meter length = 55.6 pF / meter
ql 2 o
C  ln(10000) = 9.2, C = 6.0 pf/m
V  2h  ln(1000) = 6.9, C = 8.1 pf/m Reasonable estimate is 10 pF/m
rg ln 
 r  ln(100) = 4.6, C = 12.1 pf/m
1
6. Transmission Line Models, cont.
Conductor of radius r, carrying current I
Wire over Earth
inductance
Magnetic field intensity H
h I
H  Amperes per meter .
 2 r
Surface of Earth
h Note, the image

flux exists only
above the Earth
Image conductor, at an equal distance below the Earth
 o I  h dx 2 h  r dx   o I  h2h  r   o I  2h  r 
2  r x
     ln  ln
h
x  2  rh  2  r 
 o I  2h 
 ln  for h >> r.
2  r  (4π •10‐7) / 2π Henries per meter length = 0.2 μH/m
N  o  2 h 
1 1
ln(10000) = 9.2, L = 1.8 μH/m 
L 
ext
 ln  ln(1000) = 6.9, L = 1.4 μH/m LC 1.0 106 10 1012 
I 2  r  ln(100) = 4.6, L = 0.92 μH/m
1 109 109
  
Reasonable estimate is 1 μH/m 10  1018 10 10
 3  108
2
Symmetric bundles have Double Bundle, Each Conductor Has Radius r

an equivalent radius
 
1
N 1 N A
req  NrA req  2rA
Triple Bundle, Each Conductor Has Radius r
3
req  3rA 2
A
Quadruple Bundle, Each Conductor Has Radius r
4
A
req  4rA3
3
Three phases Three Conductors Represented by Their Equivalent Line Charges
b
Dab Conductor radii ra, rb, rc
a
Dac
c
Daai Surface of Earth

Daci
ci
Dabi
ai
Images
bi
1  Daai D D 
Vag  q a ln  qb ln abi  qc ln aci 
2 o  ra Dab Dac 
Vag   p aa p ab p ac  q a 
  1 p
Vbg   2  ba pbb pbc   qb 
Vcg  o  pca pcb pcc   qc 
 
If the transmission line is symmetric, then the “P matrix” has the equal diagonal,
equal off‐diagonal property that permits 0‐1‐2 analysis rather than a‐b‐c analysis
4
 Daai Dabi Daci 

ln ln ln 
 a   ra Dab Dac   I 
a

   o ln bai D Dbbi D
 b  2  D ln ln bci   I b 
ba rb Dbc
  c    
 ln D cai D Dcci   I c 
ln cbi ln
 Dca Dcb rc 
If the transmission line is symmetric, then the “L matrix” has the equal diagonal, equal
off‐diagonal property that permits 0‐1‐2 analysis rather than a‐b‐c analysis
 LS  2 LM 0 0
Lavg
012   0 LS  LM 0 

 0 0 LS  LM 
5
Summary of Positive/Negative Sequence Capacitance and Inductance Calculations
Computation of positive/negative sequence capacitance
2 o
C /   farads per meter,
GMD / 
ln
GMRC  / 
where
GMD /   3 Dab  Dac  Dbc meters,
where Dab , Dac , Dbc are
 distances between phase conductors if the line has one conductor per phase, or
 distances between phase bundle centers if the line has symmetric phase bundles,
and where
 GMRC  /  is the actual conductor radius r (in meters) if the line has one conductor per
phase, or
N  r  A N 1 if the line has symmetric phase bundles.
N
 GMRC  /  
6
Summary of Positive/Negative Sequence Capacitance and Inductance Calculations
Computation of positive/negative sequence inductance
 o GMD / 
L /   ln henrys per meter,
2 GMRL  / 
where GMD/  is the same as for capacitance, and
 for the single conductor case, GMRL  /  is the conductor rgmr (in meters), which takes
into account both stranding and the e 1 / 4 adjustment for internal inductance. If rgmr is
not given, then assume rgmr  re 1 / 4 , and
 for bundled conductors, GMRL  /   N N  rgmr  A N 1 if the line has symmetric phase
bundles.
Computation of positive/negative sequence resistance
R is the 60Hz resistance of one conductor if the line has one conductor per phase. If the line has
symmetric phase bundles, then divide the one-conductor resistance by N.
7
Summary of Zero Sequence Capacitance and Inductance Calculations
Io → Io →
3Io → Io → 3Io → Io →
+ Io → + Io →
3Io ↓
Vo Vo
Cbundle Lbundle
Io → Io →
Lo
3Io → Io → 3Io → Io →
Lo
+ Io → + Io →
3Io ↓
Co Co Co Lo
Vo Vo
1 2 o
C0   farads per meter,
3 GMDC 0
ln
GMRC 0
where GMDC 0 is the average height (with sag factored in) of the a-b-c bundle above perfect
Earth. GMDC 0 is computed using
GMDC 0  9 D i  D i  D i  D 2 i  D 2 i  D 2 i meters,
aa bb cc ab ac bc
GMRC 0  9 GMRC3  /   D 2ab  D 2ac  D 2bc meters
8
Computation of zero sequence inductance
 
L0  3  o ln Henrys per meter,
2 GMRL 0

where skin depth   meters.
2 o f
The geometric mean bundle radius is computed using
GMRL0  9 GMRL3  /   D 2ab  D 2ac  D 2bc meters,
where GMRL /  , Dab , Dac , and Dbc were shown previously.
Computation of zero sequence resistance
There are two components of zero sequence line resistance. First, the equivalent conductor
resistance is the 60Hz resistance of one conductor if the line has one conductor per phase. If the
line has symmetric phase bundles with N conductors per bundle, then divide the one-conductor
resistance by N.
Second, the effect of resistive Earth is included by adding the following term to the conductor
resistance:
3  9.869  10 7 f ohms per meter (see Bergen),

9
As a general rule,
 C/  usually works out to be about 12 picoF per meter,

 L /  works out to be about 1 microH per meter (including internal inductance).
 C 0 is usually about 6 picoF per meter.
 L0 is usually about 2 microH per meter if the line has ground wires and typical Earth
resistivity, or about 3 microH per meter for lines without ground wires or poor Earth
resistivity.
1
The velocity of propagation, , is approximately the speed of light (3 x 108 m/s) for positive
LC
and negative sequences, and about 0.8 times that for zero sequence.
10
Ready for Use!
QL
absorbed
P1 + jQ1 P2 + jQ2
I1 R jω L I2
+ 1 1 +
200kVrms jωC/2 jω C/2 VR / δR
- -
QC1 QC2
produced produced
One circuit of the 345kV line geometry, 100km long
11
345kV Double‐Circuit Transmission Line
5.7 m
7.8 m
8.5 m
7.6 m
22.9 m at tower,
sags down 10 m
at mid‐span to
12.9 m.
7.6 m 4.4 m
22.9 m at
tower, and
Double conductor phase bundles, bundle radius = 22.9 cm,
sags down conductor radius = 1.41 cm, conductor resistance = 0.0728 Ω/km
10 m at
mid‐span to Single‐conductor ground wires, conductor radius = 0.56 cm,
12.9 m. conductor resistance = 2.87 Ω/km
12
Mack_Grady_TAMU_Relay_Conf_Tutorial_Topic7_150331.doc Mack_Grady_TAMU_Relay_Conf_Tutorial_Topic7_150331.doc
Short Circuits The symmetrical components of all a-b-c voltages are usually written in terms of the symmetrical
components of phase a by defining
1. Introduction
a  1  120 o , so that a 2  1  240 o  1  120 o , and a 3  1  360 o  10 o .
Voltage sags are due mostly to faults on either transmission systems or distribution feeders.
Transmission faults affect customers over a wide area, possibly dozens of miles, but distribution faults ~ ~ ~
Substituting into the previous equations for Va , Vb , Vc yields
usually affect only the customers on the faulted feeder or on adjacent feeders served by the same
substation transformer. ~ ~ ~ ~
Va  Va 0  Va1  Va 2 ,
Single-phase faults (i.e., line-to-ground) are the most common type of faults, followed by line-to-line, ~ ~ 2~ ~
Vb  Va 0  a Va1  aVa 2 ,
and three-phase. Since single-phase and line-to-line faults are unbalanced, their resulting sag voltages ~ ~ ~ ~
are computed using symmetrical components. Transformer connections affect the propagation of Vc  Va 0  aVa1  a 2Va 2 .
positive, negative, and zero sequence components differently. Thus, the characteristics of a voltage sag
changes as it propagates through a network. In matrix form, the above equations become
Typically, a transmission voltage sag passes through two levels of transformers before reaching a 480V ~ ~ ~ ~
Va  1 1 1  Va 0  Va 0  1 1 1  Va 
load (e.g., 138kV:12.47kV at the entrance to the facility, and 12.47kV:480V at the load). 120V loads ~   2  ~  ~  1 2  ~ 
likely have a third transformer (e.g., 480V:120V). It is not intuitively obvious how the sag changes, but Vb   1 a a  Va1  , Va1   1 a
3
a  Vb  (1)
the changes can be computed using symmetrical components and are illustrated in this report. V~c  1 a ~ ~
a 2  Va 2  Va 2  1 a
2 ~
a  Vc 
 
2. Symmetrical Components or in matrix form
An unbalanced set of N related phasors can be resolved into N systems of phasors called the symmetrical ~ ~ ~ ~
components of the original phasors. For a three-phase system (i.e. N = 3), the three sets are: Vabc  T  V012 , and V012  T 1  Vabc , (2)
where transformation matrix T is

1. Positive Sequence - three phasors, equal in magnitude, 120o apart, with the same sequence (a-b-c) as
the original phasors.
1 1 1 1 1 1
2. Negative Sequence - three phasors, equal in magnitude, 120o apart, with the opposite sequence (a-c-   1 
T  1 a 2 a  , and T 1  1 a a2  . (3)
b) of the original phasors. 3
3. Zero Sequence - three identical phasors (i.e. equal in magnitude, with no relative phase 1 a a 2  1 a 2 a 
displacement).
~ ~ ~ ~ ~ ~ ~
If Vabc represents a balanced set (i.e. Vb  Va  1  120 o  a 2Va , Vc  Va  1  120o  aVa ), then
The original set of phasors is written in terms of the symmetrical components as follows: ~ 1 ~
substituting into V012  T  Vabc yields
~ ~ ~ ~
Va  Va 0  Va1  Va 2 ,
~ ~
~ ~ ~ ~ Va 0  1 1 1   Va  0
Vb  Vb 0  Vb1  Vb 2 , ~  1 2  2 ~  ~ 
~ ~ ~ ~ 
 a1  3 1 a
V a  a Va   Va  .
Vc  Vc 0  Vc1  Vc 2 ,
V~a 2  1 a 2 ~
a   aVa   0 
 
where 0 indicates zero sequence, 1 indicates positive sequence, and 2 indicates negative sequence.
Hence, balanced voltages or currents have only positive sequence components, and the positive
The relationships among the sequence components for a-b-c are sequence components equal the corresponding phase a voltages or currents.
Positive Sequence Negative Sequence Zero Sequence
~ ~ ~ ~ ~ ~ ~ However, balanced voltages are rare during voltage sags. Most often, one phase is affected
Vb1  Va1  1  120 o Vb2  Va 2  1  120 o Va 0  Vb0  Vc 0 ~ ~ ~
~ ~ ~ ~ significantly, and the other two less significantly. Thus, all three sequence voltages Va 0 , Va1 , Va 2 exist
Vc1  Va1  1  120 o Vc 2  Va 2  1  120 o
during most sags, and these sequence voltages are shifted differently by transformers when propagating
Topic 7, Page 1 of 28- Topic 7, Page 2 of 28-

~ ~ ~
through a system. When recombined to yield phase voltages Va , Vb , Vc , it is clear that the form of phase Grounded Wye - Grounded Wye R + jX
voltages must also change as transformers are encountered.
3. Transformer Phase Shift
The conventional positive-sequence and negative-sequence model for a three-phase transformer is

shown below. Admittance y is a series equivalent for resistance and leakage reactance, tap t is the tap
(in per unit), and angle θ is the phase shift.
Bus k' y Bus k

t / :1
Bus i Grounded Wye - Ungrounded Wye
Ii ---> Ik ---> R + jX
Figure 1. Positive- and Negative-Sequence Model of Three-Phase Transformer

For grounded-wye:grounded-wye and delta:delta transformers, θ is +0˚, and thus positive- and negative-
sequence voltages and currents pass through unaltered (in per unit). However, for wye-delta and delta-
wye transformers, θ is sequence-dependent and is defined as follows:
 For positive sequence, θ is +30˚ if bus i is the high-voltage side, or –30˚ if bus i is the low-
voltage side Delta - Delta R + jX
and oppositely
 For negative sequence, θ is –30˚ if bus i is the high-voltage side, or +30˚ if bus i is the low- Figure 2. Zero-Sequence Model of Three-Phase Transformer
voltage side
It can be seen in the above figure that only the grounded-wye:grounded-wye transformer connection
In other words, positive sequence voltages and currents on the high-voltage side lead those on the low- permits the flow of zero-sequence from one side of a transformer to the other.
voltage side by 30˚. Negative sequence voltages and currents on the high-voltage side lag those on the
low-voltage side by 30˚. Thus, due to phase shift and the possible blocking of zero-sequence, transformers obviously play an
important role in unbalanced voltage sag propagation.
For zero-sequence voltages and currents, transformers do not introduce a phase shift, but they may block
zero-sequence propagation as shown in Figure 2. 4. System Impedance Matrices
Fault currents and voltage sags computations require elements of the impedance matrix Z for the study
system. While each of the three sequences has its own impedance matrix, positive- and negative-
sequence matrices are usually identical. Impedance elements are usually found by
 building the system admittance matrix Y, and then inverting it to obtain the entire Z,
or by
 using Gaussian elimination and backward substitution to obtain selected columns of Z.
The admittance matrix Y is easily built according to the following rules:

 The diagonal terms of Y contain the sum of all branch admittances connected directly to the Induced Voltage
corresponding bus. Applied Current at at Bus j
Bus k
 The off-diagonal elements of Y contain the negative sum of all branch admittances connected Power System +
directly between the corresponding busses.
V Vj
The procedure is illustrated by the three-bus example in Figure 3. Ik -
All Other Busses
ZA ZC Open Circuited
1 2 3
Figure 4. Physical Significance of the Impedance Matrix
ZE ZB ZD
I3 Impedance matrix element z j , k is defined as
Vj
z j, k  , (4)
Ik
I m  0, m 1, 2,, N , m  k
Figure 3. Three-Bus Admittance Matrix Example
Applying KCL at the three independent nodes yields the following equations for the bus voltages (with where I k is a current source attached to bus k, V j is the resulting voltage at bus j, and all busses except
respect to ground): k are open-circuited. The depth of a voltage sag at bus k is determined directly by multiplying the phase
sequence components of the fault current at bus k by the matrix elements z j , k for the corresponding
V1 V1  V2
At bus 1,  0 , phase sequences.
ZE ZA
5. Short Circuit Calculations
V2 V2  V1 V2  V3
At bus 2,   0 ,
ZB ZA ZC Short circuit calculations require positive, negative, and zero sequence impedance information,
depending on whether or the fault is balanced or not. For example, the commonly-studied, but relatively
V3 V3  V2 rare, three-phase fault is balanced. Therefore, only positive sequence impedances are required for its
At bus 3,   I3 . study.
ZD ZC
Collecting terms and writing the equations in matrix form yields Consider the three-phase fault represented by the one-line diagram in Figure 5, where VTH and Z TH
are the Thevenin equivalent circuit parameters for bus k.
 1 1 1 
   0  Zth
ZE Z A ZA  V1   0  Bus k
  1 1

1

1

1    
V2  0 ,
+
 ZA Z A Z B ZC ZC     Vth IF ZF
  V3   I 3 

 1 1 1 
0   -
 ZC Z C Z D 
or in matrix form, Figure 5. Three-Phase Fault at Bus k
YV  I , The fault current and voltage are clearly
VTH  ZF 
Besides being the key for fault calculations, the impedance matrix, Z  Y 1 , is also physically I kF  , and VkF  VTH  Z TH I kF  VTH   .
significant. Consider Figure 4. Z TH  Z F  Z TH  Z F 

V F  V Pr e   z z ka, kb z ka, kc   I ka 
F
In a large power system, the Thevenin equivalent impedance for a bus is the corresponding diagonal  kaF
  ka Pr e
  ka, ka  F 
impedance matrix element, and the Thevenin equivalent voltage is usually assumed to be 1.0 /0 pu. Vkb   Vkb    z kb, ka z kb, kb z kb, kc   I kb  ,
 F   Pr e    
Vkc  Vkc   z kc, ka z kc, kb z kc, kc   I F 
The type of machine models used when building impedance matrices affects the Thevenin equivalent  kc 
impedances and fault calculations. Rotating machines actually have time-varying impedances when
subjected to disturbances. However, for simplification purposes, their impedances are usually divided or in sequence form,
into three zones - subtransient (first few cycles), transient (5 cycles - 60 cycles), and steady-state (longer
than 60 cycles). When performing fault studies, the time period of interest is usually a few cycles, so V F  V Pr e   z 0 0 I k 0 
F
that machines are represented by their subtransient impedances when forming the impedance matrices.  kF0   kPr0 e   k 0, k 0  F 
Vk1   Vk1    0 z k1, k1 0   I k1  .
 F   Pr e    
Developing the equations for fault studies requires adept use of both a-b-c and 0-1-2 forms of the circuit Vk 2  Vk 2   0 0 z k 2, k 2   I F 
 k2 
equations. The use of sequence components implies that the system impedances (but not the system
voltages and currents) are symmetric. In general, there are six equations and six unknowns to be solved,
In abbreviated form, the above equations are
regardless of the type of fault studied.
F Pr e F
It is common in fault studies to assume that the power system is initially unloaded and that all voltages Vkabc  Vkabc  Z k  k , abc I kabc , and VkF012  VkPr012
e
 Z k  k ,012 I kF012 , (5)
are 1.0 per unit. When there are multiple sources, this assumption requires that there are no shunt
elements connected, such as loads, capacitors, etc., except for rotating machines (whose Thevenin
equivalent voltages are 1.0 pu.). where VkF consists of the voltages at bus k during the fault, VkPr e consists of the pre-fault voltages, I kF
gives the fault currents, and Z k  k contains the individual impedance elements extracted from the
Since wye-delta transformers shift positive, negative, and zero sequence components differently, it is impedance matrix.
important to model transformers according to the rules given earlier. This means that the pre-fault
voltages all have magnitude 1.0 pu., but that the pre-fault voltage angles can be 0 o ,30 o , or  30 o , The above matrix equations represents three equations (repeated in abc and 012 form), but there are six
depending upon the net transformer phase shift between them and the chosen reference bus. unknowns represented by VkF and I kF , so that three additional equations are required. The additional
equations are found by observing that
Balanced Three-Phase Fault
F F
Consider the three-phase fault at bus k, as shown in Figure 6. Vkabc  Z F I kabc , or VkF012  Z F I kF012 .
Bus k Substituting into the Thevenin equation, and recognizing that all zero- and negative-sequence voltages
a
and currents are zero for a balanced fault yields
b
 0   0   z k 0, k 0 0 0  0 
 Z I F   V Pr e    0 z k1, k1 0
 F 
c  F k1   k 1     I k1  ,
 0   0   0 0 z k 2, k 2   0 
ZF ZF ZF
so that the positive sequence fault current is found to be
IF IF IF VkPr e
kc kb ka
I kF1  1
, I kF0  0, I kF2  0 . (6)
z k1, k1  Z F
Figure 6: Three-Phase Fault at Bus k
Substituting into Thevenin equation
The Thevenin equivalent circuit equation, assuming no other current injections in the system, is
VkF012  VkPr012
e
 Z k  k ,012 I kF012 (7)

yields the fault voltage at Bus k. Similarly, because the impedance matrix relates the voltages at V F   V Pr e  0   z   I ka / 3
F
0 0
network busses to current injections at network busses, the voltage at any other bus j is found using  kF0   Prke0 Pr e
  k 0, k 0  F 
Vk1   Vk1  Vka  0 z k1, k1 0   I ka / 3
 F     
V jF012  V jPr e F Pr e
Vk 2   Vk 2  0   0 0 z k 2, k 2   I F / 3
012  Z j  k ,012 I k 012 . (8)  ka 
Note that the minus sign is needed because the fault current has been drawn as positive outward. Add the three rows yields
Once the fault voltages are known at neighboring busses, the contribution currents through the
connected branches can be easily found.
VkF0  VkF1  VkF2  Vka
F Pr e 1 F
 Vka 
 I ka z k 0, k 0  z k1, k1  z k 2, k 2 .
3

Single-Phase to Ground Fault
From the circuit it is obvious that
Consider the single-phase fault at bus k, as shown in Figure 7.
F F
Bus k Vka  I ka ZF ,
a
so that
b
c IF = 0
F
I ka Pr e 1 F
Z F  Vka 
 I ka z k 0, k 0  z k1, k1  z k 2, k 2 .
3

kb
IF = 0 F
kc ZF Solving for I ka yields
F Pr e
3Vka
I F
ka I ka  . (10)
z k 0, k 0  z k1, k1  z k 2, k 2  3Z F
Now, using
Figure 7: Single-Phase Fault at Bus k, Phase a
IF
I kF0  I kF1  I kF2  ka ,
3
As before, the Thevenin equivalent circuit equations, assuming no other current injections in the system,
is all network voltages can be found from
VkF012  VkPr012
e
 Z k  k ,012 I kF012 . V jF012  V jPr e F
012  Z j  k ,012 I k 012 .
Examining I kF012 shows that in this case Note that if z k 0, k 0  z k1, k 2 , a single-phase fault will have a higher value than does a three-phase fault.
Line-to-Line Fault
1   I ka  I F 
F
1 1   1  ka 
1
I kF012  T 1 I kabc
F
 1 a 2 F F
a   I kb  0   I ka  . (9) Consider the line-to-line fault at bus k, as shown in Figure 8.
3   3 F 
1 a 2 a   I F  0 I
 kc   ka 
Substituting into the Thevenin equation yields

Bus k
a
Using Vabc  TV012 , we find that
IF = 0
   
b ka F F
Vkb  Vkc  VkF1 a 2  a  VkF2 a  a 2 ,
c
so that
IF
IF
kc
kb
F
I kb  k1
 
V F a 2  a  VkF2 a  a 2 F
, or I kb
 
V F  VkF2 a 2  a
 k1 .
  
ZF ZF ZF
Figure 8. Line-to-Line Fault Between Phases b and c at Bus k Combining equations yields
Examining I kF012 shows that in this case  2


F  z k 2, k 2 a  a 
z k1, k1 a  a 2    I kbF Z F  V Pr e .
 a 2  a  ka
I kb 
 3 3
 
1   I ka  0 
F  
1 1 0
I kF012  T 1 I kabc
F 1
 1 a 2
a 
 F
I kb
 1 F
   I kb a  a 2  . (11) Collecting terms yields
3
1 a
2   3 F
a   I F   I F 
 I kb a 2  a
 kc kb 
 2

F  z k 2, k 2 a  a 
z k1, k1 a  a 2 ZF  
  V Pr e ,
Note that there is no zero sequence fault current.
I kb


3

3

a 2
a  


ka
Substituting into the Thevenin equation yields

or
V F   0   z  
 kF0   Pr e   k 0, k 0
0 0
 F
0
 2 


F  ZF a  a2   
  VkaPr e .
 
Vk1   Vka    0 z k1, k1 0 I
  kb a  a / 3 . I kb  z k1, k1  z k 2, k 2
 
 F  2
 a a 3 

Vk 2   0   0 0 z k 2, k 2  I F a 2  a / 3
  
 kb 
Simplifying yields
Subtracting the last two rows of the Thevenin equation yields
   
F F F
and where I kc   I kb , I ka  0 . All network voltages can now be found from
I F a  a2 I F a2  a
VkF1  VkF2  Vka
Pr e
 z k1, k1 kb  z k 2, k 2 kb ,
3 3 Pr e
F  j 3Vka
I kb  , (12)
or z k1, k1  z k 2, k 2  Z F
  2
F  z k 2, k 2 a  a
I kb 
 
z k1, k1 a  a 2   V F  V F  V Pr e . V jF012  V jPr e F
012  Z j  k ,012 I k 012 .
k1 k2 ka
 3 3 
 
From the circuit, we see that
F V F  Vkc
F
I kb  kb .
ZF

6. Calculation Procedure Step 6. While continuing to ignore the Net 30º, voltage sag propagation at all other buses j can be
computed with V jF1  V jpre F
1  Z jk ,1  I k1 .
Step 1. Pick a system MVA base and a VLL base at one point in the network. The system MVA base
will be the same everywhere. As you pass through transformers, vary the system VLL base according to
the line-to-line transformer turns ratio. Step 7. For larger networks, “hand” methods are not practical, and the Z matrix should be built. Form
the admittance matrix Y, and invert Y to obtain Z. Ignore the “Net 30º ” when forming Y. Matrix
Step 2. The system base phase angle changes by 30º each time you pass through a Y∆ (or ∆Y) inversion can be avoided if Gaussian elimination is used to find only the kth column of Z.
transformer. ANSI rules state that transformers must be labeled so that high-side positive sequence
voltages and currents lead low-side positive sequence voltages and currents by 30º. Negative sequence Step 8. Unbalanced faults also require negative sequence impedances. Faults with ground currents also
does the opposite (i.e., -30º shift). Zero sequence gets no shift. The “Net 30º ” phase shift between a require zero sequence impedances. Negative sequence impedances are usually the same as positive.
faulted bus k and a remote bus j is ignored until the last step in this procedure. Zero sequence impedances can be larger or smaller and are dramatically affected by grounding. Y∆
transformers introduce broken zero sequence paths. Prefault negative sequence and zero sequence
Step 3. Begin with the positive sequence network and balanced three-phase case. Assume that the voltages are always zero.
system is “at rest” with no currents flowing. This assumption requires that the only shunt ties are
machines which are represented as Thevenin equivalents with 1.0 pu voltage in series with subtransient Step 9. After the 012 fault currents are determined, continue to ignore the “Net 30º ” and use
impedances. Loads (except large machines), line capacitance, shunt capacitors, and shunt inductors are V jF012  V jpre F
ignored. Convert all line/transformer/source impedances to the system base using 012  Z jk ,012  I k 012 for each sequence to find 012 bus voltages.
2
 S new  V old  Step 10. Next, compute branch currents (ignoring the Net 30º) between buses j and k for each sequence
Z new old base base
pu  Z pu   old    new  . S base is three-phase MVA. Vbase is line-to-line. If a transformer
S    using
 base  Vbase 
is comprised of three identical single-phase units, Z old
pu is the impedance of any one transformer on its V jF,0  VkF,0 V jF,1  VkF,1 V jF,2  VkF,2
old I jk ,0  , I jk ,1  , I jk ,2  .
own base, and S base is three times the rated power of one transformer. For a delta connection, Vbase z0 z1 z2
line-to-line is the rated coil voltage of one transformer. For a wye connection, Vbase line-to-line is the
Step 11. As the last step, include the Net 30º between bus j and faulted bus k. Do this by adding the
rated coil voltage multiplied by 3.
Net 30º to V jF1 and I jk ,1 calculations, and subtracting the Net 30º from V jF2 and I jk ,2 calculations.
Step 4. For small networks, you can find the fault current at any bus k “by hand” by turning off all Then, use Vabc  T  V012 , I abc  T  I 012 to find the abc bus voltages and branch currents.
voltage sources and computing the positive-sequence Thevenin equivalent impedance at the faulted bus,
Z kk ,1 . Ignore the Net 30º during this step because actual impedances are not shifted when reflected
from one side to the other side of Y∆ transformers. Once the Thevenin impedance is known, then use
Vkpre
I kF1  1 , followed by VkF1  Vkpre F
1  Z kk ,1  I k1 .
Z kk ,1  Z F
Step 5. The key to finding 012 currents in a branch during the fault is to know the voltage on each end
of the branch. For a branch with positive sequence impedance z1 between busses j and k, first find
V jF1  V jpre F
1  Z jk ,1  I k1 . Positive-sequence current flow through the branch during the fault is
V jF,1  VkF,1
I Fjk ,1  . Note that z1 is the physical positive sequence impedance of the branch - it is
z1
not an element of the Z matrix. An accuracy check should be made by making sure that the sum of all
the branch currents into the faulted bus equals I kF1 .

Short Circuit Problem #1 Short Circuit Problem #2.
The positive-sequence one-line diagram for a network is shown below. Prefault voltages are all 1.0pu. A 30MVA, 12kV generator is connected to a delta - grounded wye transformer. The generator and
transformer are isolated and not connected to a “power grid.” Impedances are given on equipment
j0.1Ω Bus 2 j0.2Ω bases.
Bus 1 Bus 3
A single-phase to ground fault, with zero impedance, suddenly appears on phase a of the 69kV
transformer terminal. Find the resulting a-b-c generator currents (magnitude in amperes and phase).
j0.3Ω Regarding reference angle, assume that the pre-fault phase a voltage on the transformer’s 69kV bus has
j0.05Ω j0.1Ω angle = 0.
+ +
1/0 1/0 Gen
– – 30MVA, 12kV
Subtransient reactances Single phase to
X1 = X2 = 0.18pu Transformer ground fault
X0 = 0.12pu (Delta-GY) occurs on phase a
V j Vj Generator is connected X = 0.05pu
a. Use the definition z jk   to fill in column 1 of the Z matrix. 30MVA
I k Ik GY through a j0.5 ohm
I m  0, m  k 12kV/69kV
grounding reactor
Now, a solidly-grounded three-phase fault occurs at bus 1.
b. Compute the fault current
c. Use the fault current and Z matrix terms to compute the voltages at busses 2 and 3.
d. Find the magnitude of the current flowing in the line connecting busses 2 and 3.
1 2 3

Short Circuit Problem #3
A one-line diagram for a two-machine system is shown below. Bus1 Bus4 Bus5 Bus2
Bus6
All values in pu
on equipment
base
Bus3
The transmission line between busses 2 and 3 has X1 = X2 = 0.12pu, X0 = 0.40pu on a 100MVA,
345kV base.
Using a base of 100MVA, 345kV in the transmission line, draw one line diagrams in per unit for
positive, negative, and zero-sequences.
Then,
Use a 100 MVA, 220kv base in the transmission line.
a. Compute the phase a fault current (in pu) for a three-phase bolted fault at bus 2.
b. Compute the phase a fault current (in pu) for a line-to-ground fault at bus 2, phase a.

Short Circuit Calculations
Bus4
Short Circuit Problem #4.
Bus5
Bus1
Balanced Three-Phase Fault, Stevenson Prob. 6.15. A three-phase balanced fault, with
Bus2
ZF = 0, occurs at Bus 4. Determine
Bus6 Bus7
Bus8 Bus9
a. I 4Fa (in per unit and in amps)
b. Phasor abc line-to-neutral voltages at the terminals of Gen 1
c. Phasor abc currents flowing out of Gen 1 (in per unit and in amps)
Use 100 MVA base

Bus3
Line to Ground Fault, Stevenson Prob. 6.15. Repeat #4 for phase a-to-ground fault at
Bus 4, again with ZF = 0.

100 MVA, 138kV in the
Repeat #4, Using Stevenson Prob. 6.16.

Repeat #5, Using Stevenson Prob. 6.16.

Voltage Sag Propagation Along Feeders
Stevenson Problem 6.15, Phase A to Ground Fault at Bus #4
1. Introduction
Short circuit equations provide the theoretical framework for determining the voltage sag at a bus due to
Enter
a fault anywhere in the system. However, the short circuit equations by themselves provide little
insight. We now proceed with examples to provide this insight by showing how a sag propagates for
various transformer situations.
2. Impact of Transmission System Faults on Customers

Press
Consider the typical situation shown in Figure 1. A fault occurs at bus k of the transmission system,
causing a voltage sag that affects a substation (bus j) and the customers connected to its feeders. There
can be as many as three transformers between the customer’s load and the transmission fault point, and
Results each of these transformers can have a 30o phase shift. Typically, all three of the transformers shown
(i.e., T1, T2, and T3) are delta connected on the high side, and grounded-wye connected on the low side.
Adjacent feeder on same transformer
Transmission Substation
Feeder currents drop because of
System T1: 138kV depressed substation voltage
-----
12.47kV
12.47kV
Enter Polar Form 012 Currents at Gen #1, Compute the ABC Currents Fault T2: 12.47kV
___________________________________________________________________________ -----
480V
Customer Level 1.
Faulted bus k Substation 480V loads
transmission bus j T3: 480V
-----
120/208V
Customer Level 2.
120/208V loads
Figure 1. Example System for Analyzing the Propagation of Transmission Voltage Sags into Customer Low-Voltage Busses
The standard assumption for fault calculations is that
 the circuit is initially unloaded, or at least that the voltages are all close to 1.0 per unit.
Using this assumption, and further assuming that there are no significant contributors of fault current
on the feeders, then the actual location of the customer is not important because all points on the three
12.47kV feeders shown (including the substation 12.47kV bus) will experience the same sag.
Furthermore, the sag experienced on the substation 12.47kV bus will be the same as on substation
138kV bus j, except for possible zero-sequence component blocking and positive/negative-phase shifts.
The significance of the above paragraph is that

Enter Polar Form 012 Voltages at Gen #1, Compute the ABC Voltages

for transmission faults, one monitor at either the substation 138kV bus or at the substation 12.47kV
bus is adequate to predict voltage sag levels anywhere on the substation’s feeders, provided there are
no significant contributors of fault current on the feeders.
If the transmission fault is electrically far away, then the sag experienced at the substation and at the
customer site will be small. Alternatively, if the fault is immediately at substation 138kV bus j, then the
sag will be the most severe possible. Thus, it is reasonable to assume that an electrical “proximity”
factor exists, where a proximity factor of zero (i.e., 0%) indicates that the fault is at substation 138kV
bus j, and a proximity factor of unity (i.e., 100%) indicates that the transmission fault bus k is very far
away. From knowledge of the physical significance of the impedance matrix, and from examining
Thevenin equations
e
VkF012  VkPr012  Z k  k ,012 I kF012 ,
V jF012  V jPr e F
012  Z j  k ,012 I k 012 ,
this proximity factor P is approximated using the ratio of positive-sequence impedances
z j1, k1
P  1 . (1)
z k1, k1
By coding the short circuit equations into a Visual Basic program, and employing (1), voltage sag
propagation for the situation described in Figure 1 can now be illustrated. Assuming that the
transmission fault is relatively close to the substation (i.e., proximity factor = 25%), and that T1, T2, and
T3 are all delta:grounded-wye transformers, the line-to-neutral voltages for single-phase, phase-to- Figure 2. Propagation of Close-In Single Phase Fault on the Transmission System (all three
phase, and three-phase transmission faults are shown in Figures 2 – 4, respectively. Both phasor plots transformers have delta:grounded-wye connections)
and magnitude bar charts are given.
Note the voltage swell on phases b and c at the substation 138kV bus. Note also that two phases are
It is important to note that if a transformer is connected grounded-wye:grounded-wye or delta:delta, then affected after the first transformation, then one phase is affected after two transformations, and again
the voltage sag on the low-voltage side of the transformer is the same as on the high-voltage side, as two phases are affected after three transformations.
illustrated in Figure 5 for the single-phase fault.

Figure 3. Propagation of Close-In Phase-to-Phase Fault on the Transmission System (all three Figure 4. Propagation of Close-In Three-Phase Fault on the Transmission System (all three transformers
transformers have delta:grounded-wye connections) have delta:grounded-wye connections)

3. Impact of Distribution System Faults on Adjacent Feeders
Now, consider the situation in Figure 6 where a fault occurs on an adjacent feeder, and a monitor records
the voltage waveform at the substation 12.47kV bus.
Substation Large current Fault on adjacent

T1: 138kV Depressed current
138kV -----
12.47kV
12.47kV
T2: 12.47kV
-----
480V
Customer Level 1.
480V loads
Monitor observes T3: 480V
-----
120/208V
Customer Level 2.
120/208V loads
Figure 6. Substation Monitor Records Voltages when a Sag Occurs on an Adjacent Feeder.
As in Section 2, unless the customer’s feeder has significant contributors to the fault current, the voltage
sag at the substation 12.47kV bus will appear everywhere along the customer’s 12.47kV feeder.
However, to predict the voltage sag at Customer Levels 1 and 2, the a-b-c line-to-neutral voltages at the
substation 12.47kV bus must be
Figure 5. Situation in Figure 2 Repeated, but with all Three Transformers Having Grounded-
Wye:Grounded-Wye Connections  converted to positive/negative/zero-sequence components,
 shifted with the appropriate transformer phase shifts,
 converted back to a-b-c.

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TOPICS

VA I A  cos(2t     )  cos(2t      240)  cos(2t      240) 

cos( x)  cos( x  120)  cos( x  120)

cos( x)  cos( x) cos(120)  cos( x) cos(120),

Instantaneous Power to a Compact Fluorescent Lamp

Turns ratio 7200:240

Open Circuit Test

Φ Open circuit test: Open circuit

X / R Ratios for Three-Phase Transformers

%Load loss = 0.9

%No-load loss = 0.2 7200V 240V

Modern Distribution Transformer:

A three-phase transformer can be three separate single-phase

Wye-Equivalent One-Line Model

• Reflect to side 2 using individual transformer

N1:N2 Standard 345/138kV autotransformers,GY-GY ,

For Modeling a Delta-Delta Connection, Convert the Transformer to

Wye-Equivalent One-Line Model

N1:N2 • Convert side 1 impedances from delta to

For Modeling a Delta-Wye Connection, Convert the Transformer to

Wye-Equivalent One-Line Model

For Modeling a Wye-Delta Connection, Convert the Transformer to

Wye-Equivalent One-Line Model

N1:N2 • Reflect to side 2 using three-phase bank line-to-line

From conservation of power,

2 Moving the impedances from low-

2 Moving the impedances from low-

• Calculate Ibase(rms) = Sbase / Vbase. Zbase = Vbase / Ibase = (Vbase)2 / Sbase.

• Zpu = Zactual / Zbase, so Zactual = Zpu • Zbase

• Zactual = Zpu_old • Zbase_old. Also, the same Zactual = Zpu_new • Zbase_new

• Thus, Zpu_new = Zpu_old • Zbase_old / Zbase_new

S base ,1 S base ,3 S base ,3

Z  V 2 baseLL ,3 ,old   S base ,3 ,new 

Grounded Wye - Delta R + jX

Grounded Wye - Ungrounded Wye

Ungrounded Wye - Delta R + jX

Trans1 TLine1 TLine2 Trans3

Gen1 Trans1 Line1 Line2 Trans3 Motor

For generators and transformers,

Fortescue's Theorem: An unbalanced set of N related phasors can be resolved into N

• Nega ve Sequence (“−” or “2”) ‐ three phasors, equal in magnitude, 120o apart,

• Zero Sequence (“0”) ‐ three identical phasors (i.e. equal in magnitude, with no

Grounded Wye - Delta R + jX

Grounded Wye - Ungrounded Wye R + jX

Ungrounded Wye - Delta R + jX

Z 012  T 1Z abcT

Image conductor, at an equal distance below

h Note, the image

Image conductor, at an equal distance below the Earth

Symmetric bundles have Double Bundle, Each Conductor Has Radius r

Triple Bundle, Each Conductor Has Radius r

Quadruple Bundle, Each Conductor Has Radius r

Daai Surface of Earth

 Daai Dabi Daci 

Computation of positive/negative sequence capacitance

GMD /   3 Dab  Dac  Dbc meters,

where Dab , Dac , Dbc are

Computation of positive/negative sequence inductance

where GMD/  is the same as for capacitance, and