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POWER BASIC
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Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
1
Copyright SEL 2005
Power System Protection for Engineers
Power System BasicsOverview
The purpose of this introductory section is to provide a review of the main concepts of power systems, especially those concepts most frequently used in the power system protection field.
It is difficult to attempt to study all aspects of power systems in a short period of time. This section serves not only as a brief overview but also as an introduction to conventions used throughout the course.
The concepts presented here can also be found in any textbook on ac circuit analysis or power system analysis.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
2
Power System BasicsObjectives
z Explain phasor concepts and their importance in power systems
z Discuss the voltage, current and power relationships in three-phase electric circuits
z Review the power system components and per-unit quantities
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
3
Sinusoidal Function
2 =period
mY
( )ty
t
t
2
2
mY
( ) ( ) += tYty m cos
The current and voltage of alternating current electrical systems in steady state are normally represented by perfect sinusoidal functions. The figure shows an example of a sinusoidal signal (or function) called y(t) that could be a voltage or a current. The signal is periodic with a period of T seconds. The signal frequency f, in Hertz, is the reciprocal of the period: f = 1/T
The analytical expression of a sinusoidal function of period T is the following:
( ) ( )
locationon depending Hz, 60or 50ff2
anglePhase:frequency(angular)Radian :
AmplitudePeak :Y
tcosYty
m
m
= =
+=
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
4
Complex Number
Real Axis
Imaginary Axis
Yr
)YIm(b
)YRe(a
jbaY
==
+=
a
b
The analysis of ac linear systems using the time representation of voltages and currents may result in tedious, complicated, and time-consuming computations. For steady state conditions, the complication is reduced through the use of complex numbers. A complex number is composed of two real numbers. One of the real numbers is the real part and the other real number is the imaginary part. The real part of the complex number Y shown in the figure is a, and the imaginary part is b. The imaginary part is always multiplied by j, which is equal to the square root of negative one.
A complex number can be graphically represented as shown in the figure. Two axes are used to represent the real and imaginary parts of the number. The number can be represented as a vector with two components. The following three expressions can be used to represent a complex number:
= =+=
+==
+=
sinYbcosYa
:are partsimaginary and real theand,baYis Y of magnitude theSince
angle) and magnitude using formar (Rectangul sinjYcosYYform) coordinate(Polar YY
form) coordinatear (RectanguljbaY
22
rrrr
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
5
Phasor Representation
( ) [ ]tj
jjm
eYRe2ty
:Then
Yee2
YY
:asY Phasor the Define
=
==
r
r
r
A phasor is a complex number used to represent an ac voltage or current. The relationship between the phasor and the original signal is given by the following analytical development:
( ) ( ) ( )
( ) ( ) ( )[ ]( ) ( )[ ] [ ]
1j
sinjcose
ee2
YRe2)t(y
eeYReeYRety
tsinjYtcosYRety
tcos2
Y2tcosYty
j
tjjm
tjjm
tjm
mm
mm
=
+=
===
+++=
+=+=
+
:Note
:IdentitysEuler' (*)
:(*)identity sEuler' Using
:is time of function a as signal original The
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
6
Phasor Representation
( )tyofvalueRmsYYWherejYYY
YYeY
m
j
:2
sincos
=
+===
r
r
Note that the phasors magnitude Y, is the rms value of the original sinusoidal signal y(t). Note also that the phasor real and imaginary parts, as well as thephasors magnitude and angle are CONSTANT. In other words, with this representation the variable t (time) does not appear in the calculations.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
7
Notation
( )
+=
== ++
:ThereforeA"Phasor of Angle" Means
/ )(
A
A
AAeA j
r
r
r
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
8
Phasor Representation
Real Axis
Imaginary Axis
= YYrsinY
cosY
As any complex number, a phasor can be graphically represented in the complex plane.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
9
Phasor Operations
2*
)(/*
)(/
:tionMultiplica/
/
:Given
AAAABBA
ABBA
BB
AA
==
+=
==
rrrrrr
rr
Phasor operations are the same as for complex numbers. Multiplication of phasors is easiest when they are in polar form.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
10
Phasor Operations
nj
nn
jnnnjn
j
eA
eAAeABAe
BA
BA
=====
A
)()( :tionExponentia
)(/ :Division )(
Other operations, like division and exponentiation, of complex number are relatively simple to perform. This would not be so simple if the time representation of the signals was used.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
11
Representation of AC Linear CircuitsUsing Phasors and Impedances
V
I
IV Z
IMPEDANCE
( )
REACTANCEXRESISTANCER
jZZjXRZ
ZI
VI
VIVZ IV
I
V
==
+=+===
==
sincos
/rrr
Linear elements of passive ac circuits are represented by their impedances. The impedance of a given circuit element is defined as the complex number resulting from the division of the applied voltage phasor by the resulting current phasor. As with any complex number, the impedance has a real and an imaginary part. The real part is called RESISTANCE (R) and the imaginary part is called REACTANCE (X).Examples of impedances are:
Z = 2 + j6 Ohms Z = 2 - j30 OhmsZ = j10 Ohms (pure reactance)Z = 2.5 Ohms (pure resistance)Z = 0.231 + j0.685 per unit (using a given base impedance)Z = 23.1 + j68.5 % (per unit times 100)
The inverse of impedance is called ADMITTANCE and denoted by Y. For some applications in power systems, the admittance is used instead of the impedance to represent passive elements. The admittance also has real and imaginary parts. The real part is called CONDUCTANCE (G) and the imaginary part is called SUSCEPTANCE (B). In other words:
jBGjXR
1Z1Y +=+== r
r
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
12
Passive Circuit Element Impedances
I
+ V -
Resistor
R I
+ V -
LI
+ V -
C
Inductor Capacitor
VI
I I
VV
= 0 = 90 = - 90
Even though pure elements do not exist, it is common to model some elements as pure resistors, inductors, and capacitors. The phasor diagrams for these three elements are shown in the figure.
The following expressions are used to calculate the impedances of the three main components of linear ac circuits:
cc
LL
jXC
1jZ,IjXIC
1jV:Capacitor
jXLjZ,IjXILjV:Inductor
RIVZ,IRV:sistorRe
========
===
rrrrrrr
rrrrr
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
13
I
V
VL VCVR
R L C
VRI
VC
VRVL
VC
V
Series RLC Circuit Phasor Diagram
The following equations serve to calculate the total series impedance of the series R-L-C circuit:
( ) == C1LjXXjX CLC1jLjRjXjXRZ
CL +=+=
jXRIVZ +==
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
14
Parallel RLC Circuit Phasor Diagram
V
IR
IICIL
IC
I
R L CIR IL IC
IR
V
The parallel R-L-C circuit is the dual of the series R-L-C circuit. It is more convenient in this case to use admittances.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
15
Impedance Representation in the Complex Plane
R
X
Z
R
X
r
-r
r-r
ArearZ || r
22 || XRZZ
jXRZ
+==+=
rr
As with any complex number, an impedance can be represented in the complex plane. In the power system protection field, the complex plane used to represent impedances is called the R-X plane. The resulting diagram is sometimes called the R-X diagram.
The R-X diagram is used to study and analyze not only a simple impedance (which is actually a simple point on the complex plane) but also a set of impedances or a variation of impedances. For example, the area inside the circle of radius r and with center at the origin is represented by:
rZ || r
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
16
Instantaneous Power
)2sin()sin()2cos()cos()cos(
)2cos()cos(
)cos()cos(2
)cos(2
)cos(2
tVItVIVIp
tVIVIp
ttVIivp
tIi
tVv
++=+=
===
= Zi
v
The single-phase ac circuit of the figure serves to provide a review of the concept of instantaneous power for steady state conditions. Note that, for this particular case, it is assumed that the current lags the voltage by an angle of degrees. Note also that the magnitude of each signal is presented as the rms value times the square root of two.
The instantaneous power is obtained by direct multiplication of the two sinusoidal functions representing the voltage and the current. The result is a function with three terms:
1) The first term is constant (does not depend on t), and is equal to VIcos()2) The second term is a perfect sinusoid but at a frequency equal to 2. The magnitude of this term is proportional to the cosine of angle .3) The third term is a perfect sinusoid also at a frequency equal to 2. The magnitude of this term is proportional to the sine of angle .
Note that, if the impedance of the circuit were a perfect resistor, the current and the voltage would be in phase. In other words, the angle would be zero. Note also that the mathematical expression of the instantaneous power can be re-written as:
)t2sin(sinVI))t2cos(1(cosVIp)t2sin(sinVI)t2cos(cosVIcosVIp
++=++=
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
17
Instantaneous Power
( ) += tVIVIp 2coscosp = vi
ivP
)cos( 10
VIdtpT
PT
Average ==The slide shows how the mathematical expression of the instantaneous power can be manipulated to make its form more evident. The final expression has a constant term (VIcos()) and a double-frequency sinusoidal term. The instantaneous power is shown in the figure as the offset sinusoidal function with an offset equal to the constant term VIcos(), which is shown as the dashed line in the figure. Intuitively, it can be seen that the constant offset is the average power delivered to the impedance.
The figure also shows the voltage and current signals.
The average value can be found by integrating the expression and the result is, as expected, VIcos().
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
18
Average Power of Elements
I
+ V -
Resistor
R I
+ V -
LI
+ V -
C
Inductor Capacitor
= 0 = 90 = -90tVIVIp 2cos+= tVIp 2sin= tVIp 2sin=
Average = VI= I 2 R
Average = 0 Average = 0
The average power delivered to pure inductors and capacitors is zero. It is said that resistors are the only elements that consume real power (active power).
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
19
Complex PowerActive and Reactive
=
=
IVReIVReP **
=
=
IVImIVImQ **
jQPIVS * +== rrrI
VZ
Complex power is defined as the product of the voltage phasor and the complex conjugate of the current phasor. The unit of measure for complex power is volt-amperes, or VA.
The average power is also known as the active power, because it is the part of the instantaneous power that actually produces work, or heat. The unit of measure for active power is watts. Another way to determine the active power consists of taking the real part of the multiplication of the voltage and current phasors.
The imaginary part of the complex power, Q, is known as the reactive power. The unit of measure for reactive power is Volt-Amperes-Reactive, or VAR.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
20
Complex PowerActive and Reactive
=
= cosVIIVReP
=
= sinVIIVImQ
IV
Z
+=+= sinjZcosZjXRZr
=+== VIjQPIVS *rr
If is the angle of the impedance, and therefore the angle the current lags the voltage, then the active and reactive power can be calculated as functions of .Note that, according to convention, the reactive power of an inductor is positive and the reactive power of a capacitor is negative.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
21
Complex PowerApparent Power and Power Factor
( ))FactorPower(cos
SPPF
PowerApparentZIQPSVIS
)PowerComplex(IVjQPSeS
222
*j
==
=+===
=+== r
rrr
IV Z
Also by definition, the magnitude of the complex power (the product of the magnitudes of the voltage and current, S = VI) is called the apparent power.
The ratio of the active power to the apparent power is defined as the power factor. Mathematically, the power factor is the cosine of angle , or p.f. = cos().
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
22
Power Triangle
P
SQ
It is possible to graphically represent the complex power in the complex plane. The real part is the active power and the imaginary part is the reactive power. This is known as the power triangle.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
23
Balanced Three-Phase Systems
z For a balanced three-phase system, the sinusoidal voltages are the same amplitude, displaced in phase by 120.
VA
VB
VC
VA
VC
VB
A-B-C Sequence A-C-B Sequence
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
24
Y- Connected Loads
EBnECn
EAnn
ZG
ZG
VCn ZG VBn
Ib
Vbn
Zp
n
Zp
Vcn
Zp
VanIa
ZLVAn
ZL
Ic
ZL
ooo 1201200 ===
PCNPBNPAN VVVVVV
The figure shows a circuit diagram of a simple three-phase system. If the ideal three-phase voltage source is perfectly balanced, and all the impedances on each phase of the system are equal, then all voltages and currents in the system will be perfectly balanced.
A wye-connected load consisting of passive impedances is shown for thepurposes of reviewing the equations.
Each impedance of the load receives a line-to-neutral voltage.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
25
Y- Connected Loads
o303 ==
PBNANABVVVV
o903 ==
PCNBNBCVVVV
o1503 ==
PANCNCAVVVV
o30V3V PLL =
30
VCNVCA VAB
VAN
VBC
VBN
Each of the impedances of a load is called a phase impedance. The magnitude of the voltage applied to each impedance is the line-to-neutral voltage, or the phase voltage (VP).
The relationship between the line-to-line voltages and the the line-to-neutral voltages can be obtained analytically or graphically. There is a factor equal to the square root of three between the phase voltage and the line-to-line voltage.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
26
Y- Connected Loads
==
/
P
P
ANA I
Z
VI
==
o120/P
P
BNB I
Z
VI
==
o120/P
P
CNC I
Z
VI
PL II =
The magnitude of the current passing through each of the impedances is the line current (IL). For a Y-connected load, the line current and the phase currents are equal.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
27
- Connected Loads
PLL VV =
Ica
IabZp
Ibc
Iaa
Ibb
IccZp
Zp
For a delta-connected load, the load voltage is equal to the line-to-line voltage.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
28
- Connected Loadsoo 120II,120II,0II PCAPBCPAB ===
o1503 ==
PABBCB IIII
o903 ==
PBCCAC IIII
o303 =
PLII
30
IC
ICA
IAB
IAIBCIB
o303 ==
PCAABAIIII
For delta-connected loads, there is a factor equal to the square root of three between the phase and the line currents.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
29
Conclusion for Passive Loads
LPLPII,3VV ==
3II,VVLPLP
==
z Y- Connected Loads
z - Connected Loads
This is the conclusion.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
30
Power in Balanced Three-Phase Systems
== cosIV3cosIV3PLLPP
== sinIV3sinIV3QLLPP
22 QPIV3IV3SLLPP
+===QjPeSS j +==
== cosS/PPF
Three-phase power is obtained by adding up the power on each phase. The result is that the three-phase power for a balanced three-phase system is three times the power of one of the phases. When the line-to-line voltages and the line currents are used, the factor becomes the square root of three.
Recall that for wye-connected loads, VP = VL/ 3 and IP = IL. When the substitution is made, the equation has 3 divided by root 3, resulting in root 3 times the line voltage and line current.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
31
Conclusion on Balanced Power Systems
EBnECn
EAn
n
ZG
ZG
VCn ZG VBn
Ib
Vbn
Zp
n
Zp
Vcn
Zp
Van
Ia
ZLVAn
ZL
Ic
ZL
If the system is perfectly balanced, only one phase needs analysis. The other phases will have the same magnitude with a 120 phase shift.
Ean
ZG Zp
ZL
Ia
+-
One important characteristic of the analysis of three-phase power systems is that it is not necessary to analyze each phase when the system is perfectly balanced. For a balanced system, only one phase needs to be analyzed. The behavior of the other phases is similar. The only difference is that the currents and voltages are shifted 120 with respect to the analyzed phase.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
32
One-Line Representation of Power Systems
EBnECn
EAnn
ZG
ZG
VCn ZG VBn
Ib
Vbn
Zp
n
Zp
Vcn
Zp
VanIa
ZLVAn
ZL
Ic
ZL
SystemRepresentation
One-LineRepresentation
The one-line representation saves significant amounts of space. This is more evident for large power systems.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
33
Some Power System ComponentsOne-Line Symbols
Generator
Transformer
Transmission Line
General Load Shunt Impedance
These are the symbols used in one-line diagrams for the most common elements of a power system. Other elements are capacitor banks, reactors, phase shifters, motors, power electronics controls, circuit breakers, etc.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
34
Power System ExampleOne-Line Diagram
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
35
In the Substation Yard
Bus
SwitchCircuit
Breaker CurrentTransformer
VoltageTransformer
Switch
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
36
Mutual CouplingAn Important Concept
MagneticCoupling
InducedVoltages
I1 I2 I1 I2Zm I2 Zm I1
Magnetic coupling between two energized circuits is called mutual coupling. The current in one of the circuits induces a voltage in the other circuit and vice-versa.
Mutual coupling is present in all the components of the three-phase power system. This makes the study, analysis, and computation more complicated.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
37
Power System ComponentsPower Generator
ZSIa
Ib
Ic
g
n
Bus
Stator
ZS
ZS
ZmZm
Zm
Ze
Va+
-Vb+
-Vc+
-
VrIr
Ires
Rotor (Field) Circuit
Ia
Ib
Ic
NeutralGroundingDevice
The synchronous generator is a complex machine. Windings in the rotor inject direct current, but the windings move. The movement of the windings is what produces the induced ac voltages in the stator circuits. The stator windings and the rotor windings are magnetically coupled and the equivalent mutual inductances are variable because the machine is in motion. The mathematical model of a typical synchronous generator consists of at least 10 differential equations, including the control. There are simplified models used in practical calculations that provide enough accuracy for many applications.
The generators neutral can be grounded with different methods. In the model shown in the figure, the grounding impedance can take on different characteristics depending upon the method used.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
38
Simplified Generator Model
ZS
+Ea
+Eb
+Ec
Ia
Ib
Ic
g
n-
-
-
Bus
ZS
ZS
ZmZm
Zm
Ze
Va+
-Vb+
-Vc+
-
gnccSbmamc
gnbcmbSamb
gnacmbmaSa
VVIZIZIZE
VVIZIZIZE
VVIZIZIZE
++++=++++=++++=
The most common, and simple, method to represent a generator in steady state is with an ideal three-phase source with series and symmetrically coupled impedances.The internal ac sources represent the induced voltages, that is the rotor effect. The equation considers the self impedance, Zs, of each winding of the stator windings and the mutual impedances, Zm, equal for all cases (perfect symmetry). When the system works under perfectly balanced conditions, the ground-to-neutral current and voltage Vgn are both zero.
Although simple and inaccurate, the described model is useful for practical studies and is especially useful for creating equivalent three-phase generators to represent the power system elements behind a given bus. This is the three-phase version of a Thvenin equivalent.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
39
Power System ComponentsPower Lines and Cables
(Not to scale)
Overhead lines and cables are among the most important elements of the power system. They transport and distribute the energy along vast geographical regions. There is a large variety of line configurations (conductor arrangement, tower design, etc.). However, the lines can be modeled with an acceptable degree of accuracy with relatively simple models.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
40
Magnetic Coupling
a b c
MagneticCoupling
Long lines and cables have mutual coupling among their own phases and among other adjacent lines and cables. This makes the analysis of three-phase power systems a complicated task, specifically for unbalanced conditions. Some special techniques, such as the symmetrical components method, are used to address this complication.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
41
Transmission Line Circuit Model
IaIb
Ic
a
b
c
Magnetic Coupling
One of the simplest models for transmission lines consists of the circuit shown in the figure. Both the inductive series impedance and the shunt capacitance are actually distributed along the entire line. These parameters are concentrated in this model to simplify the study of the system in steady state conditions.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
42
Transmission Line Circuit ModelWithout Considering Capacitances
IaIb
Ic
a
b
c
Magnetic Coupling
aV bV cV aV bV cV
When studying power system short circuits using phasors, the line capacitance effect can be ignored, resulting in an even simpler line model. Note the mutual coupling representation.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
43
Line Equations Without Considering Mutual Coupling Among Phases
ccSc
bbSb
aaSa
VIZVVIZVVIZV
+=+=+=
Ib
Ic
a
b
c
aV bV cV aV bV cV
If the mutual effect is neglected, the line model is too inaccurate. This model is sometimes used for extremely short lines.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
44
Line Equations Considering Mutual Coupling Among Phases
ccSbcbacac
bcbcbSabab
acacbabaSa
VIZIZIZVVIZIZIZVVIZIZIZV
+++=+++=+++=
IaIb
Ic
a
b
c
Magnetic Coupling
aV bV cV aV bV cV
The steady state equations of the line, neglecting the capacitance and considering the mutual coupling, are the ones shown in the figure. In this case, the effect of the grounding wire is not shown. This effect can be easily considered by adding a similar coupled equation for each ground wire on the line.
The equations presented in the slide take into account that, in real lines, the self impedances (Zs) are similar for all phases, but the mutual impedances are not all equal. There is some symmetry. For example, Zac=Zca, Zbc=Zcb, etc. These equations are known as the equations for a non-transposed line.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
45
Line Equations for a Symmetrical (Transposed) Line
ccSbmamc
bcmbSamb
acmbmaSa
VIZIZIZVVIZIZIZVVIZIZIZV
+++=+++=+++=
IaIb
Ic
a
b
c
Magnetic Coupling
aV bV cV aV bV cV
If the line is considered perfectly symmetrical, the mutual impedances are all equal. This model is known as the transposed line model, and is widely used because the inaccuracy of the model is evident only for very specific applications.
The self impedance has two parts, resistance and reactance. The resistance depends on the conductor used (cross section, material, etc.). The reactance depends on the conductor type and the geometric position of the conductors. The mutual impedance is purely reactive and depends almost exclusively on conductor arrangement and characteristic. The soil characteristics and the grounding wires also have a remarkable influence on the line impedances.
Examples of line impedances for a 13 kV line, with 4/0 ACSR conductor:
Zs = 0.3272 + j1.07 Ohm/km = 0.524 + j1.721 Ohm/mile
Zm = j0.636 Ohm/km = j1.018 Ohm/mile
These values were obtained assuming that the line is perfectly symmetrical.
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
46
Power System ComponentsTransformer
z Single Phasez Three Phase
Power System Basics_r16
Section 1 - Power System Basics
Power System Protection for Engineers PROT 401
47
Single-Phase Transformer
Real Transformer: N1I1 = N2I2+ ERROR
N1 N2
I1 I2
ZLV1
The figure shows the most common representation of a single-phase transformer. The flux confined to the iron core produces the coupling between the two windings. The leakage flux, together with the resistance of the windings and the core non-linearity, produce an error that is negligible in many applications.
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Single Phase Transformer Symbol
N1 N2
I1V2
+
-
V1
+
-
I2
2
1
2
1
NN
VVTR
N
N =Transformer Ratio:
The transformer ratio is the ratio between the rated voltages for the two terminals of the transformer.
In a well-designed transformer, the ratio between the rated voltages is very close to the turns ratio.
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Section 1 - Power System Basics
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Ideal Single Phase Transformer
N1 N2
I1V2
+
-
V1
+
-
I2
221121
2211
2
1
2
1
IVIVSSININ
NN
VV
=====
Perfect relationships can only be applied in the so-called ideal transformer. Even though this device does not exist in reality, the model is used as an auxiliary for more complex models in power systems.
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Section 1 - Power System Basics
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Impedance Reflection
2
2
2
1
2
2
1
2
2
1
22
2
12
1
11
2
22
ZNN
NN
IV
NNI
NNV
IVZ
IVZ
=
===
=
N1 N2
I1V2
+
-V1+
-
I2Z2
I1V1+
-(N1/N2)2 Z2
The ideal relationships described in the previous slide lead to the fact that an impedance connected to one side of the ideal transformer is reflected to the other side as the same impedance times the square of the transformer turns ratio.
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Section 1 - Power System Basics
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Single Phase TransformerEquivalent Circuit
I2
N1 N2
IE
Rs jXs
Non-Linear
Ideal
Rp jXp
I1
V1 V2+
-
+
-
Usually Not Considered
The figure shows the most popular model for a single-phase transformer.
The equivalent circuit of a single-phase transformer includes the following elements:
An ideal transformer, to represent the effect of the main flux
Leakage reactances, to represent the transformer leakage flux and other effects.
The series resistances, to represent the winding ohmic losses.
The shunt magnetization, a non-linear impedance, is used to represent the behavior of the magnetic circuit (iron core). This branch is usually not considered, especially for short-circuit studies.
There are more accurate models, but this one provides enough accuracy for most practical applications.
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Section 1 - Power System Basics
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Single Phase TransformerSimplified Equivalent Circuit
Rs jXs
Ideal
Rp jXp
2
1
2
1
NN
VV
x
x =
I2
N1 N2
I1V1 V2
+
-
+
-
V1x
+
-
V2x
+
-
This is how the equivalent circuit looks after disregarding the magnetization branch. Note that for the two internal (and imaginary voltages V1x and V2y) the ideal transformer relationship still holds. Then, the winding impedances can be reflected to one side or the other, depending on the specific application.
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Section 1 - Power System Basics
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Per Unit Representation
z Quantity in Per Unit
z Actual Quantity & Base Value Scalar or complex value of power,
voltage, current, or impedance
z Quantity in Percent Quantity in Per Unit x 100
Actual QuantityBase Value of Quantity
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Section 1 - Power System Basics
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Selection of Base Quantities for Single-Phase System
bV
bS
Choose Voltage Base :
Choose Power Base :
The Current Base and Impedance Base are Derived from the Chosen Values:
b
b
b
bb
b
bb S
VIVZ
VSI
2
; ===
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Section 1 - Power System Basics
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Effect of Per Unit Calculations on a Transformers Equivalent Circuit
z Step 1: ChooseVoltage Base Vb1Power Base Sb
Rs jXs
Ideal
Rp jXp
V2I2I1
V1+
-
+
-
V1x+
-
V2x+
-
2
1
2
1
NN
VV
x
x =z Step 2: Choose
Voltage Base Vb2 = (N2/N1)Vb1
When using Per Unit values, the power base must remain constant. Therefore, the power base is selected once and then used for all further calculations.
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Section 1 - Power System Basics
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Effect of Per Unit Calculations on a Transformers Equivalent Circuit
I2(pu)
Rs(pu) jXs(pu)Rp(pu) jXp(pu)
V1(pu) V2(pu)+
-
+
-V1x(pu)
+
-
V2x(pu)+
-
I1(pu)
disappears rtransforme ideal The=
====
1)pu(V)pu(V
NN
VVand
NN
VV
;VV)pu(V;VV)pu(V
x2
x1
2
1
b2
b1
2
1
x2
x1
2bx2x21bx1x1
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Section 1 - Power System Basics
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Effect of Per Unit Calculations on a Transformers Equivalent Circuit
I2(pu)
Rs(pu) jXs(pu)Rp(pu) jXp(pu)
V1(pu) V2(pu)+
-
+
-
V1x(pu)=V2x(pu)+
-
I1(pu)
I(pu)V1(pu) V2(pu)+
-
+
-
ttt jXRZ +=
Conclusion: If calculations are performed using per unit values, a proper choice of the base causes the transformer to disappear. This is the main advantage of the per unit method.
The impedance of power transformers normally comes on the nameplate of a transformer and is given as a percentage (Z (pu) x 100) of the rated impedance base for the transformer.
As can be seen from the per unit equivalent circuit, the impedance of a transformer can be measured by making a short circuit test of the transformer. This is because of the following:
Zt = V1(Per Unit)/I1(Per Unit) with V2 = 0 (short circuit)
This is why the transformer impedance is usually referred to as the short circuit impedance.
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Section 1 - Power System Basics
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Example
z Single-Phase Distribution Transformer TR = 7200/120 V/V SN = 100 kVA Zt = 3%
Manufacturer provides the transformers series impedance in percent (100 x Z pu) using the transformers rating as the power base.
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Section 1 - Power System Basics
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Equivalent Circuit in Per Unit
I(pu)V1(pu) V2(pu)+
-
+
-
tt jXZ =
I(pu)V1(pu) V2(pu)+
-
+
-
pu03.0jZt =
This is the equivalent circuit for the transformer of the previous slide.
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Three-Phase Banks
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Y-Y Transformer
N1 N2 Ia
Ib
Ic
Ia(N2/N1)
Ib(N2/N1)
Ic(N2/N1)
a
b
c
2
1
2
1
NN
KVKVTR =
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Section 1 - Power System Basics
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-Y Transformer
N1 N2 Ia
Ib
Ic
(Ia-Ib)(N2/N1)
(Ib-Ic)(N2/N1)
(Ic-Ia)(N2/N1)
a
b
c
2
1
2
1
3NN
KVKVTR =
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Section 1 - Power System Basics
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Per-Unit System Advantages
z Gives Clear Idea of Relative Values of Similar Quantities
z Per-Unit Impedance of Equipment Based on Rating Fall in Narrow Range
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Section 1 - Power System Basics
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Per-Unit System Advantages
z Per-Unit Transformer Values Same Regardless of Transformer Connection or Reference to Primary or Secondary Side
z Per-Unit System Ideal For Computer-Based Power System Analysis and Simulation
Note: The phase shift is considered in per-unit representation. This is done in posterior analysis. The magnitudes, however, are the same in per-unit.
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Section 1 - Power System Basics
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Three-Phase Base Quantitiesz Typically Select
Sbase (Three-Phase Value) Vbase (Line-to-Line Value)
base
basebase
base
basebase kV3
kVAI
V3SI ==
( )base
2base
base
basebase S
VI
3VZ ==
( ) ( )base
2base
base
2base
base MVAkV
kVAkV1000Z ==
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Section 1 - Power System Basics
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Change of Base
( )2oldbase
old
baseold
base
old
pu
VSZZ
ZZ ==
( )2newbasenewbase
newbase
newpu
VSZZ
ZZ ==
2
newbase
oldbase
oldbase
newbaseold
punewpu V
VSSZZ
=
It is often necessary to convert an impedance to a different base. For example, if a transformer impedance is given on a 20 MVA base, it may need to be converted to a 100 MVA base to match the base of a system model. If the voltage base is not being changed, that portion of the equation becomes equal to 1 and can be ignored.
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Example
Generator Transformer
z Find Equivalent Reactance on 100 MVA, 115 kV Base Generator: 50 MVA, 13.2 kV, XG = 15% Transformer: 50 MVA, 13.8/115 kV,
XT = 8%
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Section 1 - Power System Basics
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Example
pu4345.0XXX
pu16.050
10008.0X
pu2745.08.132.13
5010015.0X
TG
T
2
G
=+==
=
=
=
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Section 1 - Power System Basics
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Example of Power Flow Through a 69kV Transmission Line
V1 V2
jQPSI
+= LLL jQPS +=
z Sending-End Voltage (Line-to-Neutral)and Three-Phase Power:
z Load Voltage (Line-to-Neutral)and Three-Phase Power:
)MVAR,MW(45j60S;kV06.37V L2 +==
)MVAR,MW(26.58j65.62S;kV05.689.42V1 +==
12.5 miles
This is an example of steady state functioning of a power line. The voltages are phase-to-neutral voltages, and the powers are three-phase. It is more common to present the voltages in line-to-line volts (for normal load) or in per-unit (or percent) of a given base.
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Section 1 - Power System Basics
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Example of Power Flow Througha 69 kV Transmission Line in Per Unit
jQPSI
+= LLL jQPS +=
pu45.0j60.0S;pu09438.0V L2 +==
pu583.0j627.0S;pu05.6077.1V1 +==
V1 V2Base:69 kV100 MVA
12.5 miles
z Sending-End Voltage (Line-to-Neutral)and Three-Phase Power:
z Load Voltage (Line-to-Neutral)and Three-Phase Power: