Power System
Calculations—Part
Calculations
Part II
Kurt B. Ederhoff, Ph.D., P.E.
2010 Industry Applications Society Annual Meeting - Houston, TX
October 3-7, 2010
2
I
Introduction
d i
• Topics
▫ Per Unit Basis
▫ Symmetrical
S
t i lC
Components
t
▫ Component Modelling
▫ Faults
l & Sequence Networks
k
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P Unit
Per
U i C
Conversion
i
• Per Unit Definition
A quantity a (voltage, current, power, impedance,
admittance,, etc.)) in per
p unit is defined as the ratio
of that quantity to a selected base quantity of the
same nature (i.e. voltage, current, power,
i
impedance,
d
admittance,
d itt
etc.)
t )
a[per unit]
POWER SYSTEM CALCULATIONS—PART II
[units]
a[actual units]
a
= [actual units]  a[pu] = [units]
ab
ab
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P Unit
Per
U i
• Example:
If V = 120 [[V]] and we select Vb = 80 [[V],
], then
V
[pu]
POWER SYSTEM CALCULATIONS—PART II
120 [V]
=
= 1.5 [pu]
80 [V]
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6
P Unit
Per
U i
• Percent Definition
If the per unit value is multiplied by 100, the
quantityy is expressed
q
p
in p
percent with respect
p
to
the base quantity.
For the
F
th previous
i
example,
l we say the
th voltage
lt
iis
150% (1.5 x 100) of the base voltage of 80 [V].
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P Unit
Per
U i
• Per unit conversion requires us to select a base
quantity
• How do we make the selection?
▫ Answer: Select two quantities as the base from the
following: voltage, current, power, impedance,
admittance
• Which do we choose?
▫ Answer:
A
Generally
G
ll choose
h
voltage
lt
and
d power.
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P Unit
Per
U i
• Why Voltage and Power?
▫ Voltage. For each voltage level in our system, we
know the rated voltage of equipment, and even if
loading changes,
changes the voltage does not deviate too much
from the rated value.
▫ Power. The range of power flowing in a section of the
system is quadratically related with the voltage
voltage. As
such, the range of expected power flow is known for an
area. Note, for transmission level analysis, it is
c sto a to select a base power
customary
o e of 100 MVA
MVA.
Note: The base power is usually selected to be the
same for the entire network.
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C
Conventions
i
• Unless specifically stated,
stated the following
quantities are always complex numbers:
V,, I,, Z,, Y,, S,, t
• All other quantities are real numbers, such as
R, X, G, B, L, C, P, Q
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T
Terminology
i l
V
I
Z
Y
S
t
R
X
G
B
Voltage
Current
Impedance
Admittance
Complex Power
Turn Ratio
Resistance
Reactance
Conductance
Susceptance
POWER SYSTEM CALCULATIONS—PART II
L
C
P
Q
|S|
Inductance
Capacitance
Active Power
Reactive Power
Apparent Power
• Pet Peeve #1: There is no
“real power" anywhere
here
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P Unit
Per
U i
• Selecting Quantities
Vb = Vf -f
Sb = S3f
▫ Voltage
phase-to-phase
p
voltage
g at each
 Usuallyy selected as the nominal p
voltage level
▫ Power
g of 3
3-phase
p
p
power flowing
g in the
 Usuallyy selected in the range
network (i.e. whatever network is being analyzed)
 For transmission level analysis, it is customary to select a base
power of 100 MVA.
 The base power is usually selected to be the same for the entire
network.
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P Unit
Per
U i F
Formulas
l
S3[VA]
f b
I f[A]-n b =
Z
[W ]
b
=
POWER SYSTEM CALCULATIONS—PART II
[V]
f -f b
3V
(V
)
2
[V]
f -f b
[VA]
3 f
f b
S
,
,
Vf[V]
- n b =
[Siemens]
b
Y
Vf[V]
-f b
3
1
= [W]
Zb
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P Unit
Per
U i ffor E
Equipment
i
• Per unit quantities for equipment is defined at
the equipment level
▫ Ex:
25
5 MVA,, 110-33
0 33 kV transformer
a so e
Z = 8.3%
▫ Ex:
4 kV,
k 1500 hp,
h 0.88 PF, 93.5% eff
ff
Xd” = 0.155 pu
Smotor
1500 [hp] ´ 0.746 [hp/kW]
= Sb =
0.88 PF ´ 0.935 eff
= 1360 [kVA]
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P Unit
Per
U i C
Conversion
i
• Equipment ratings vary
• Selection of base quantities vary
• Analysis requires use of a common base
A[pu.old base]
[units]
A[units]
A
= [units] ; A[pu.new base] = [units]
Ab.old
Ab.new
[units]
[pu.new base] [units]
A[pu.old base] Ab.old
=
A
Abb.new
b ld
A[pu.new base] = A[pu.old base]
POWER SYSTEM CALCULATIONS—PART II
[units]
Ab.old
b old
[units]
Ab.new
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P Unit
Per
U i C
Conversion
i
• Particularly for impedances
Remember:
[V] 2
(Vb )
[W ]
Zb =
Sb[VA]
[W ]
æ
[pu.new base]
[pu.old base] ç Z b.old
Z
= Z
çç [W]
èZ
ö÷
÷÷
÷
b.new ø
[V]
æ
V
= Z [pu.old base] ççç b.old
[V]
V
è
ö÷
÷÷÷
b.new ø
POWER SYSTEM CALCULATIONS—PART II
2
[VA] ö
æ Sb.new
÷÷
çç
çè S [VA] ÷ø÷
b.old
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Ad
Advantages
Of P
Per Unit
U i
• Equipment Parameters. For a given type of equipment, and
disregarding the size and voltage, the parameters in per unit are
within a narrow, known range
j
networks of different
• Eliminate Turn Ratio. For two adjacent
voltage levels, if the selected base power is the same throughout and
the selected base voltages match the turn ratio of the transformer
between the networks, then all quantities in per unit have the same
value regardless of which voltage level they are defined. In essence,
the transformer is eliminated.
• Eliminate Coefficients. For almost all equations with quantities
defined in per unit, the numerical coefficients are eliminated.
• Voltage. In per unit, the line-to-neutral voltage equals the phase-tophase voltage, and during normal operation both quantities are
close
l
to unity.
i
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History
Hi
• Charles LeGeyt Fortescue
▫ 1876 – Born in York Factory, Manitoba
▫ 1898 – First engineer to graduate from Queens
University at Kingston in Ontario
▫ Joined Westinghouse after graduation and spent his
entire career there
▫ 1913 – Co-authored
h d paper on measurement off hi
high
h
voltage using sphere gap, a method still used to this
day (97 years later)
▫ Obtained 185 patents in his career in design
transformers, insulators, and DC and AC power
circuits
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History
Hi
• Charles LeGeyt Fortescue (cont
(cont’d)
d)
▫ 1921 – Elected AIEE Fellow
▫ 1930 – Paper in Electric Journal that outlined
“direct stroke theory”, which is said to have
completely revolutionized the approach to the
li h i problem.
lightning
bl
L
Led
d to adoption
d i off overhead
h d
static lines.
▫ 1936 – Died in December at age 60
POWER SYSTEM CALCULATIONS—PART II
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History
Hi
• Charles LeGeyt Fortescue (cont
(cont’d)
d)
▫ 1918 - Transactions of the American Institute of
Electrical Engineers (AIEE) included the classic paper
y
Co-Ordinates Applied
pp
to the
Method off Symmetrical
Solution of Polyphase Networks
88 pages long
24 additional pages of discussion
303 numbered
b d equations
i
“bewildering exhibit of subscripts to be found in it is
something that will, well, make one pause”
 Vladimir Karapetoff suggested that the term
“symmetrical components” was a “more correct and
descriptive” expression




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S
Symmetrical
i l “C
“Co-Ordinates”
O di
”
• What are Symmetrical Components?
▫ Any set of N unbalanced phasors — that is, any such
“polyphase” signal — can be expressed as the sum of N
symmetrical sets of balanced phasors.
phasors
▫ Only a single frequency component is represented by
the phasors. This is overcome by using techniques
such as Fourier or LaPlace transforms.
transforms
▫ Absolutely general and rigorous and can be applied to
both steady state and transient problems.
▫ It is thoroughly
h
hl established
bl h d as preeminently
l the
h only
l
effective method of analyzing general polyphase
network problems
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22
Th
Three-Phase
Ph
SSequence N
Networks
k
• Three
Three-Phase
Phase Systems
▫ Three sets of symmetrical components, where each set
is referred to as a sequence.
phasors,, called the p
positive sequence,
q
, has
▫ First set of p
the same phase sequence as the system under study
(say ABC)
▫ The second set, the negative sequence, has the reverse
phase
h
sequence ((BAC)
AC)
▫ The third set, the zero sequence, phasors A, B and C
are in phase with each other.
▫ Method
M th d converts
t any sett off th
three phasors
h
i t th
into
three
sets of symmetrical phasors, which makes asymmetric
analysis more tractable.
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S
Symmetrical
i lC
Components
• Set of three phasors,
phasors say Xa, Xb and Xc can be
represented as a sum of the three sequence sets
X a = X a 0 + X a1 + X a 2
X b = X b 0 + X b1 + X b 2
X c = X c 0 + X c1 + X c 2
where
X a 0 , X b 0 , X c 0 is the zero sequence
q
set
X a1 , X b1 , X c1 is the positive sequence set
X a 2 , X b 2 , X c 2 is the negative sequence set
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S
Symmetrical
i lC
Components
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S
Symmetrical
i lC
Components
Only three of the sequence values are unique,
unique
X a , X a1 , X a 2
The others can be determined. First,, we define a
complex operator a.
a  e j120 = 1120
a 2 = e j 240 = 1240
3
j 360
a = e
= 1360 = 1
1 + a + a2 = 0
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S
Symmetrical
i lC
Components
Then, we relate phases within a sequence to the reference,
which is a-phase in this example
X a0 : X b0 = X a0 ,
X c0 = X a0
X a1 : X b1 = a 2 X a1 X c 1 = aX a1
X a 2 : X b 2 = aX a 2
X c2 = a2 X a2
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S
Symmetrical
i lC
Components
Rearrange the equations
X a = X a 0 + X a1 + X a 2
2
X b = X a 0 + a X a1 + aX a 2
X c = X a 0 + aX a1 + a 2 X a 2
which in matrix form looks like this:
é1 1 1 ù é X a 0 ù
éXa ù
ú
ê
úê
ê ú
2
a ú ê X a1 ú
ê X b ú = ê1 a
ú
ê
úê
êX ú
2
êë1 a a úû ëê X a 2 ûú
êë c úû
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S
Symmetrical
i lC
Components
Using simple notation, we can convert from symmetrical
component reference to the phase reference as follows:
X phase = TX sym
where
h
é1 1 1 ù
ê
ú
2
T  ê1 a
aú
ê
ú
2
êë1 a a úû
Of course,
course the following equation can convert from the
phase reference to the symmetrical component reference:
X sym = T -1 X phase
h
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S
Symmetrical
i lC
Components
The inverse of the transformation matrix is
é1 1 1 ù
ú
1ê
-1
2
T = ê1 a a ú
ú
3ê
2
êë1 a
a ûú
which in detail, results in the following equations:
é1 1 1 ù é X a ù
é X a0 ù
úê ú
ê
ú
1ê
2
ê X a1 ú = ê1 a a ú ê X b ú
úê ú
êX ú
3ê
2
êë1 a
a úû ëê X c ûú
êë a 2 úû
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S
Sym.
Components
C
– Example
E
l 1
Let
Then
éI a ù
é 100 ù
ê ú
ê
ú
I = ê I b ú = ê10  ú
êI ú
ê 10 ú
êë c úû
êë
úû
I s = T -1 I
é1 1
1ê
= ê1 a
3ê
êë1 a 2
POWER SYSTEM CALCULATIONS—PART II
é  ù
1 ù é 100 ù
ú
ú
ê
ú
2 ê
a ú ê10  ú = ê100ú
úê
ú
ê
ú
a úû êë 10 úû
êë 0 úû
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S
Sym.
Components
C
– Example
E
l 2
If
Then
POWER SYSTEM CALCULATIONS—PART II
é 100 ù
ê
ú
I = ê10  
ú
ê10  ú
êë
úû
é 0 ù
ê
ú
Is = ê 0 ú
ê100ú
êë
úû
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S
Sym.
Components
C
– Example
E
l 3
If
Then
POWER SYSTEM CALCULATIONS—PART II
éVa
ê
êVb
ê
êVc
ë
ù
é 8.00 ù
ú
ê
ú
ú = ê 6.0
6 0 - 90 ú
ú
ê
ú


16.0
143.1
ú
êë
úû
û
éV0
ê
êV1
ê
ê
ëV2
ù
é 2.0143.1 ù
ú
ê
ú
ú = ê 9.8
9 818
18.44 ú
ú
ê
ú


4.3
86.2
ú
êë
úû
û
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33
S
Symmetrical
i lC
Components
POWER SYSTEM CALCULATIONS—PART II
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S
Symmetrical
i lC
Components
POWER SYSTEM CALCULATIONS—PART II
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S
Symmetrical
i lC
Components
POWER SYSTEM CALCULATIONS—PART II
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O Way
One
W to Vi
Visualize
li
• Positive Sequence
▫ Can be thought of as a “source” that rotates a
machine in a p
particular direction
• Negative Sequence
▫ This “source” rotates a machine in a direction
opposite of that of the positive sequence
• Zero Sequence
▫C
Causes no rotational
t ti
l fforce, but
b t instead
i t d fforms an
oscillating (not rotating) field in the machine
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C
Comments
on SSym. C
Components
• The transformations apply only for linear systems,
that is, systems with constant parameters
(impedance, admittance) independent of voltages
and currents
• The quantities used for X can be phase-to-neutral or
phase-to-phase voltages, or line or line-to-line
currents.
• For some connections, the zero sequence component
is always zero
▫ Line currents for an ungrounded wye connection
▫ Line currents for a delta connection
▫ Line-to-line voltages for a delta connection
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M
More
C
Comments
• Symmetrical components is a important tool for
analyzing unbalanced states
▫ Three-phase unsymmetrical network and state is
converted to three symmetrical networks and states
▫ Symmetrical networks can be solved using single
phase techniques
▫ With
Wi h symmetrical
i l components we solve
l three
h
interconnected symmetrical networks using single
phase analysis, which is easy.
▫ Once solved, we use transformation equations to
obtain phase quantities
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M
More
C
Comments
• Equipment Parameters
▫ Symmetrical components has advantage that
parameters in system
p
y
components
p
are easier to
define
▫ Because each sequence is a symmetrical threephase
h
case, the
h parameters can b
be d
defined
fi d using
i
typical three-phase tests.
POWER SYSTEM CALCULATIONS—PART II
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40
Complex Power with Symmetrical
Components
For wye connections
*
S = V I + V I +V I + Vn ( I a + I b + I c )
*
an a
*
bn b
*
n
a
*
b b
*
cn c
= (Van + V ) I + (Vbn + Vn ) I b* + (Vcn + Vn ) I c*
= Va I a* + V I + Vc I c*
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Complex Power with Symmetrical
Components
For delta connections
S = Vab I ab* + Vbc I bc* +Vca I ca*
= (Va - Vb ) I ab* + (Vb - Vc ) I bc* + (Vc - Va ) I ca*
*
*
*
= Va ( I ab - I ca ) + Vb ( I bc - I ab ) + Vc ( I ca - I bc )
= Va I a* + Vb I b* + Vc I c*
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Complex Power with Symmetrical
Components
In matrix form
*
éI a ù
ê ú
T
S = éëVa Vb Vc ùû ê I b ú = Vphase
I*phase
ê ú
êIc ú
ë û
We have the following transforms
Vphase = TVsym
I phase = TI sym
Then the complex power becomes
T
S = Vsysym
T T T * I*sysym
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Complex Power with Symmetrical
Components
The product
é1 0 0ù
ê
ú
T *
T T = 3I identity = 3 ê0 1 0ú
ê0 0 1ú
êë
úû
Then the complex power becomes
T *
S = 3 Vsym
I sym
= 3(Va 0 I a*0 + Va1 I a*1 + Va 2 I a*2 )
If the voltages and current are defined in
per unit,, the "3" disappears
p
pp
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U off SSymmetrical
Use
i lC
Components
• Consider the following wye connected load
I n = I a + Ib + I c
Va = ZY I a + Z n I n
Va = ( ZY + Z n )I a + Z n I b + Z n I c
Vb = Z n I a + ( ZY + Z n )I b + Z n I c
Vc = Z n I a + Z n I b + ( ZY + Z n )I c
éVa ù
é ZY + Z n
ê ú
ê
êVb ú = ê Z n
êV ú
ê Z
êë c úû
êë
n
POWER SYSTEM CALCULATIONS—PART II
Zn
ZY + Z n
Zn
ù éI a ù
Zn
úê ú
Zn
ú ê Ib ú
ZY + Z n úûú êëê I c úûú
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U off SSymmetrical
Use
i lC
Components
éVa ù
é ZY + Z n
ê ú
ê
ZY
êVb ú = ê Z n
êV ú
ê Z
êë c úû
êë
n
V = ZI
V = TVsym
I = TI sym
TVsym = ZTI sym
Vsym = T -1ZTI sym
é Z y + 3Z n
ê
-1
T ZT = ê
0
ê
0
êë
POWER SYSTEM CALCULATIONS—PART II
Zn
+ Zn
Zn
0
Zy
0
ù éI a ù
Zn
úê ú
Zn
ú ê Ib ú
ZY + Z n úúû êêë I c úúû
0 ùú
0ú
ú
Z y úû
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U off SSymmetrical
Use
i lC
Components
éV0 ù
é ZY + 3 Z n 0
ê ú
ê
êV1 ú = ê
0
ZY
ê ú
ê
0
0
êV2 ú
ê
ë
ë û
Systems are decoupled
V0 = ( Z Y + 3 Z n ) I 0
V1 = ZY I1
V2 = Z Y I 2
POWER SYSTEM CALCULATIONS—PART II
0 ù éê I 0 ùú
ú
0 ú ê I1 ú
ê ú
ú
ZY úû ê I 2 ú
ë û
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Generators
• Fault Behavior
▫ Sudden change in voltage and current, such as
those in faults,, produces
p
transients
▫ Armature current divided into two components
 Symmetrical AC component – whose associated
component in
i the
h fi
field
ld is
i a DC current
 DC component – whose associated component in the
field is an AC current
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Generators
• Symmetrical Component Modelling
▫ Principle concern is with symmetrical component and its
associated constants
▫ DC component often eliminated from studies
 Usually not necessary to apply or set protective relays
 If necessary (e.g. circuit breaker applications), various factors
are available from standards,, manufacturers,, or other sources
▫ For synchronous machines, symmetrical AC component can
be resolved into three distinct components
 Subtransient component – the double prime (“) values
 Transient component – the single prime (‘) values
 The steady-state component
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Generators
• Subtransient Component
▫ Occurs during onset of fault
▫ Subtransient reactance (Xd”)) approaches armature
leakage reactance but is higher as a result of
damper windings, and so on.
▫ Subtransient time constant (Td”) is very low
(because damper windings have relatively high
resistance) typically around 0
resistance),
0.01
01–0
0.05
05 seconds
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Generators
• Transient Component
▫ Armature current demagnetizes the field and decrease
flux linkages with the field winding
▫ Transient reactance ((Xd’)) includes effect of both
armature and field leakages and is higher than
armature leakage reactance, and thus higher than the
subtransient reactance
▫ Transient
i
time
i
constant ((Td’) varies
i typically
i ll ffrom 0.35
to 3.3 seconds
• Steady-State Component
▫ Transient
i
eventually
ll d
decays
▫ For faults, eventually becomes unsaturated direct axis
reactance (Xd)
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Generators
POWER SYSTEM CALCULATIONS—PART II
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Generators
• Negative Sequence
▫ Subtransient reactance can be measured by blocking
the rotor with the field winding shorted and applying
single phase voltage across any two terminals
▫ As position of rotor is changed, measured reactance
varies considerably if machine has salient poles
without dampers (and very little damper winding
exists) or if the machine has a round rotor
▫ For negative sequence, similar phenomenon exists
exceptt rotor
t iis att 2ff with
ith relation
l ti tto fi
field
ld sett up b
by
applied voltage
▫ Good approximation: X 2 = 12 X d¢¢ + X q¢¢
(
POWER SYSTEM CALCULATIONS—PART II
)
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58
Generators
• Zero Sequence
▫ Varies quite a lot
▫ Depends largely on pitch and breadth factors of
armature winding
▫ Generally, X0 is much smaller than X1 and X2
values
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59
Generators
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G
Generators
&N
Network
kE
Equivalents
i l
• “Strong”
Strong Remote
Systems
▫ Assume X1 = X2
▫ Calculate X1 from 3phase fault duty
▫ Calculate
C l l t X0 from
f
SLG fault duty
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I d i M
Induction
Machines
hi
• Positive Sequence
▫ Changes from stalled to
running
▫ ~0.15
5p
pu stalled ((Xd”))
▫ 0.9–1.0 pu running
• Negative Sequence
▫ Remains
e a s eeffectively
ect e y co
constant
sta t
▫ ~0.15 pu (Xd”)
• Zero Sequence
▫ 0.0 if wye ungrounded or
delta connected
62
T
Transformers
f
• Modelling
▫ Usually modelled as a series impedance
▫ Shunt parameters can be calculated by review of
transformer tests.
▫ Shunt parameters don’t generally impact analysis
▫ Transformer winding configuration determines
sequence networks
▫ Three winding transformers have interesting
sequence networks, but close inspections shows
them to be intuitive
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T
Transformer
f
Sequence
S
Networks
N
k
64
3 Winding
Wi di T
Transformer
f
IImpedance
d
• 3 Winding Transformer Impedance
▫ Usually given as a winding-to-winding (delta)
impedances in per cent
q
wye
y impedances
p
for sequence
q
▫ Convert to equivalent
network analysis
▫ Often times, the base power is different for various
impedances
▫ Ex:
100 MVA auto with 35 MVA tertiary
May show ZHM on 100 MVA base
May show ZHL and ZML on 35 MVA base
Must convert delta impedance to common base
before converting to equivalent wye network
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65
3 Winding
Wi di T
Transformer
f
IImpedance
d
• Delta-Wye
Delta Wye Conversion Formula
Z H = 12 ( Z HM + Z HL - Z ML )
Z M = 12 ( Z HM + Z ML - Z HL )
Z L = 12 ( Z HL + Z ML - Z HM )
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66
Zi Z Transformer
Zig-Zag
T
f
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T
Transmission
i i Lines
Li
• Positive and Negative Sequence Impedance
▫ Passive component
▫ Assume line transpositions
m0 Dm
L1 = L2 =
ln
[H/m]
2p Ds
X = w L = 2p fL
Dm
-7
X 1 = X 2 = 4p ´ 10 ln
[W/m]
Ds
R1 = R2 = conductor AC resistance (tables)
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68
T
Transmission
i i Lines
Li
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69
T
Transmission
i i Lines
Li
• Zero Sequence Impedance
▫ More involved
▫ Assumptions
 Zero sequence current divides equally between
conductors
 Conductors
C d
are parallel
ll l to ground
d
 Earth is a solid with a plane surface, infinite in
extent,, and of uniform conductivityy
▫ None of the assumptions are true
▫ We get acceptable error with these assumptions
▫ Line design affects calculation techniques
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70
T
Transmission
i i Li
Line Z
Zero SSequence
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71
T
Transmission
i i Li
Line Z
Zero SSequence
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72
T
Transmission
i i Li
Line Z
Zero SSequence
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73
T
Transmission
i i Li
Line Z
Zero SSequence
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T
Transmission
i i Li
Line Z
Zero SSequence
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75
Wh We
What
W Won’t
W ’ C
Cover
• What We Won’t
Won t Cover
▫ Capacitive Reactance of Transmission Lines
▫ Mutual Impedance of Transmission Lines
▫ Cable
C bl Modelling
M d lli (look
(l k up iin ttable
bl or contact
t t
manufacturer)
• So What’s the General Idea?
▫ Produce a positive, negative, and zero sequence model
for your components.
▫ Simple model for equipment is a series impedance
▫ Connect the models according to system topology
▫ Analyze imbalances by connecting systems
POWER SYSTEM CALCULATIONS—PART II
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77
F l
Faults
• Definition
▫ A fault is any condition in a system considered
abnormal
yp
▫ Basic Types
 Short-Circuit Faults
 Open-Circuit Faults
 Combined Faults
• Shunt Faults
▫
▫
▫
▫
Three-Phase (3Ph)
Phase-To-Phase (LL)
( )
Phase-To-Phase-To-Ground (LLG)
Single-Line-To-Ground (SLG)
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78
F l
Faults
• Series Faults
▫
▫
▫
▫
Open phase
Open neutral
Two open phases
Impedance in a phase
• Combination Faults
▫ See connection diagrams
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Sequence Connections – Shunt
Faults
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Sequence Connections – Shunt
Faults
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Sequence Connections – Series
Faults
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Sequence Connections – Series
Faults
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Sequence Connections –
Combination Faults
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84
E
Example
l 1 – System
S
Parameters
P
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E
Example
l 1 – System
S
Parameters
P
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E
Example
l 1 – System
S
Parameters
P
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Example 2 – Delta Wye Transformer
3-Ph Fault
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Example 2 – Delta Wye Transformer
3-Ph Fault
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Example 3 – Delta Wye Transformer
SLG Fault
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Example 3 – Delta Wye Transformer
SLG Fault
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Example 4 – Delta Wye Transformer
LL Fault
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Example 4 – Delta Wye Transformer
LL Fault
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D l W Summary
Delta-Wye
S
• 3
3-Phase Fault
▫ Behaves as expected, same
current both sides
• LL Fault
▫ Must coordinate full fault
current on primary with
0.866 factor on secondary
(i.e. coordination interval)
• SLG Fault
F lt
▫ For solidly grounded
system, shift damage curve
577 to p
protect against
g
byy 0.577
full fault current on
secondary
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E
Example
l 5 – Resistor
R i
Eff
Effect
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E
Example
l 6 – Open
O
Ph
Phase F
Fault
l
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E
Example
l 6 – Open
O
Ph
Phase F
Fault
l
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