DRAFT and INCOMPLETE
Table of Contents
from
A. P. Sakis Meliopoulos
Power System Modeling, Analysis and Control
Chapter 4 _____________________________________________________________ 2
Modeling - Power Transformers and the Per Unit System ______________________ 2
4.1 Introduction____________________________________________________________ 2
4.2 Single Phase Transformers________________________________________________ 2
4.2.1 The Ideal Single Phase Transformer _____________________________________________ 3
4.2.2 Impedance Transformation ____________________________________________________ 5
4.2.3 Theory of Mutually Coupled Inductances _________________________________________ 7
4.3 The Per Unit System ____________________________________________________ 14
4.3.1 Axiomatic Definition of the Per-Unit Scaling System _______________________________ 16
4.3.2 Per-Unit Scaling Procedure ___________________________________________________ 16
4.3.3 Mutually Coupled Circuits ____________________________________________________ 21
4.3.4 The Per Unit System and Numerical Accuracy ____________________________________ 26
4.3.5 Discussion_________________________________________________________________ 28
4.4 Three Phase Transformers_______________________________________________ 29
4.4.1 The Ideal Three Phase Transformer _____________________________________________ 31
4.4.2 Non-Ideal Three Phase Transformer Model _______________________________________ 33
4.4.3 Sequence Circuits of Three Phase Transformers ___________________________________ 34
4.5 Autotransformers_______________________________________________________ 39
4.5.1 Single-Phase Autotransformers _____________________________________________ 39
4.5.2 Three Phase Autotransformers _____________________________________________ 41
4.6 The Regulating Transformer _____________________________________________ 41
4.7 Transformer Power Flow Equations _______________________________________ 47
4.8 Transformer Parameter Identification From Tests ___________________________ 51
4.8.1 The Open Circuit Test________________________________________________________ 51
4.8.2 The Short Circuit Test________________________________________________________ 52
4.9 Normalized Power System Model _________________________________________ 53
4.10 Summary and Discussion _______________________________________________ 55
4.11 Problems ____________________________________________________________ 56
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Chapter 4
Modeling - Power Transformers and the Per
Unit System
4.1 Introduction
In this chapter we examine the power transformer. Single phase, three phase as well as
regulating transformers are considered. We start from the physical construction of a
transformer, we examine the operation of the transformer and develop a mathematical
model which naturally leads to an equivalent circuit. This approach is followed for all
types of transformers, single phase, three phase or regulating transformers. Transformer
models can be further simplified with the use of specific scaling of electrical quantities,
otherwise known as the per unit system. The per unit system has been extensively used
in modeling and analysis of power systems with multiple transformation levels of voltage
(transformers). It is natural to introduce the per unit system in this chapter.
4.2 Single Phase Transformers
A single phase power transformer consists of a magnetic core and two windings as it is
illustrated in Figure 4.1a. This is a rather simple construction of a magnetic circuit. The
excitation of the magnetic circuit is a result of electric current flow in the two windings.
Specifically when electric current flows in one or both windings, a magnetic flux will be
generated inside the iron core of the transformer. The alternating magnetic flux induces
voltages on the two windings and it is responsible for the power transfer from one
winding to another. The two windings many times are referred to as primary and
secondary, which typically indicate the source side and the load side, respectively.
Another description of the two windings is high voltage side or low voltage side
depending on the voltage level at the two windings.
When the electric current flows in one winding of the transformer of Figure 4.1, a
magnetic field will be generated in the core of the transformer. If the electric current is
alternating, the magnetic field will be also alternating. Consider the other coil. The
alternating magnetic flux links this coil. As a result, a voltage will be induced in this
coil. The polarity of the induced voltage with respect to the current i1(t) will depend on
how the coil is wound. We can define the polarity of the transformer, and we symbolize
the polarity with two dots as it is illustrated in Figure 4.1a, as follows. If the direction of
the electric current flow in a coil is “into” the dot, then the induced voltage on the other
coil will have positive polarity on the dot side of this coil. Considering the construction
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
of the transformer of Figure 4.1a, an electric current i1 flowing “into” the dot, will
generate a magnetic flux in the indicated direction (use the right hand rule). This
magnetic flux will induce a voltage on coil 2 which will have a positive polarity on the
top terminal. This discussion should explain how the dots have been placed. Now, using
the dots we can forget the construction of the transformer and instead use the symbolic
representation of Figure 4.1b.
Φm
i1
i'2
Φl2
Φ l1
(a)
i1
i'2
(b)
Figure 4.1 Single Phase Transformer
(a) Construction
(b) Symbolic Representation
4.2.1 The Ideal Single Phase Transformer
Power transformers are typically constructed with high conductivity wires (copper), and
therefore low resistance. The magnetic core is typically made from carbon steel which
has a very high permeability and it is laminated to minimize losses. It is expedient to
take advantage of these observations, for the purpose of examining the basic properties of
transformers. Specifically, we can idealize the parameters of the transformer, yielding
the concept of the “ideal transformer”. The specific assumptions of the “ideal
transformer” are
•
•
No losses ( in the windings or in the core)
Core material has infinite permeability, i.e. µ = ∞ which implies that there is no
leakage fluxes
The “ magnetic resistance” of the magnetic circuit of Figure 4.1a is:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
1 l
,
µ A
where l is the length of the magnetic circuit, A is the cross section of the core, and µ is
the permeability of the core material. For the “ideal transformer”, ℜ = 0 , and the total
magnetomotive force for the transformer of Figure 4.1 is
ℜ=
P = N1i1 (t ) + N 2i2 (t ) = ℜϕ (t ) = 0
Therefore:
i1 (t )
N
=− 2
i 2 (t )
N1
Note that the magnetiuc flux f(t) is linked by both windings. The magnetic flux linkage of
the two windings are λ1 (t ) = N1ϕ (t ) and λ2 (t ) = N 2ϕ (t ) respectively. Therefore the
following induced voltages will appear on the two windings:
e1 (t ) =
dλ1 (t )
dϕ (t )
= N1
,
dt
dt
e2 (t ) =
dλ2 (t )
dϕ (t )
= N2
dt
dt
Therefore (upon elimination of the flux variable):
e1 (t ) N 1
=
e 2 (t ) N 2
Above relationship indicates that the voltages at the terminals of an ideal transformer are
inversely proportional to the number of turns of the two windings.
The total power absorbed by the ideal transformer is
p (t ) = e1 (t )i1 (t ) + e 2 (t )i 2 (t ) =
⎛ N ⎞
N1
e 2 (t )⎜⎜ − 2 ⎟⎟i 2 (t ) + e 2 (t )i 2 (t ) = 0
N2
⎝ N1 ⎠
Above relationships are valid for the instantaneous values of voltage, current and power.
Under sinusoidal steady state conditions, these relationships are also valid for the phasors
of the voltage and current and the complex power, i.e.
~
E1 N 1
~ =
E2 N 2
~
I1
N2
~ =−
N1
I2
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
~~
~~
S1 = E1 I1* = − E 2 I 2*
The above relationships characterize an ideal transformer. Practical transformers are
designed so that their operation under normal operating conditions is close to the ideal
case.
4.2.2 Impedance Transformation
A transformer inserted between a source and an impedance (load) can transform the
apparent impedance seen by the source as it is illustrated in Figure 4.2. If the transformer
is near ideal, this transformation of the impedance can be achieved without any
substantial power losses in the transformer. Considering Figure 4.2, the impedance at the
left-hand side of the transformer is given with:
~
I2
~
I1
~
I1
~
V2
Source
z2 ⇔
2
Source
⎛N ⎞
z1 = ⎜⎜ 1 ⎟⎟ z2
⎝ N2 ⎠
N1:N2
Figure 4.2 Impedance Transformation by a Transformer
N1 ~
V2
2 ~
2
~
⎛ N 1 ⎞ V2 ⎛ N 1 ⎞
V1 N 2
⎟
⎟ Z
Z1 = ~ =
=⎜
=⎜
N 2 ~ ⎜⎝ N 2 ⎟⎠ I~2 ⎜⎝ N 2 ⎟⎠ 2
I1
I2
N1
Note that the insertion of a transformer with transformation ratio N1/N2 between a source
and a load of impedance Z2, resulted in an apparent impedance (N1/N2)2Z2.
This property is many times used for ‘’matching’’ a load to a specific source. Consider
the problem of connecting a source to a load and that the source has an internal
impedance of ZS while the load has an impedance ZL. For what value of the load
impedance ZL, the source will transfer maximum power to the load? The power absorbed
by ZL will be
(
)
~~
P = Re V L I L* = Re(Z L )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
~
E
Zs + ZL
2
=
RL
~2
E
( RS + RL ) 2 + ( X S + X L ) 2
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
where
~
VL is the voltage at the terminal of the load,
~
E is the internal voltage source.
Z s = Rs + jX s
Z L = R L + jX L
Now we would like to find the maximum of P by allowing the load impedance ZL to
change. The maximum can be computed by solving the equations.
∂P
=0
∂ RL
∂P
=0
∂ XL
The solution of above equations yields:
R L = RS
X L = −X S
or
Z L = Z S*
Many times it is possible to achieve the above conditions (or come very close to it) by
inserting a transformer between the source and the load. An example will illustrate this
application.
Example E4.1: Consider a source with a 50 ohm internal impedance. The source is to
be connected to an 8 ohm load as it is illustrated in Figure E4.1. How can we transfer the
maximum possible power to the load.
50Ω
8Ω
100Vrms
Amplifier
Speaker
Figure E4.1 A Source and a Load
Solution: If the load is directly connected to the source, the power transferred to the load
will be computed as follows. First the electric current is
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
I = 100/58 = 1.724 A (rms)
The power absorbed by the load is
P = Re(VLI*) = Re[(8)(1.724)2] = 23.78 Watts
For maximum power transfer, the load must be connected to the source via a transformer
in such a way that the impedance of the transformer/speaker system, as indicated in
Figure E4.1a, be: Z ′L = Z*S
This means that (N1/N2)2⋅8Ω = 50 Ω and therefore N1/N2= 2.5. In this case
I' =100V/100Ω = 1 A, and I = 2.5 I ′ = 2.5A. The maximum power absorbed at the load
is P = (8Ω)(2.5A )2 = 50Waltts .
50Ω
I'
Z'L
I
8Ω
100Vrms
N1:N2
Figure E4.1a
4.2.3 Theory of Mutually Coupled Inductances
In this section we consider a single phase transformer without the “ideal” assumption, i.e.
the windings of the transformer have finite resistance and the magnetic core has finite
permeability. In this case, the magnetic flux generated by the flow of currents in the
windings is not confined in the magnetic core but some magnetic flux “leaks” in the air
as it is illustrated in Figure 4.3. Specifically Figure 4.3a illustrates the magnetic flux due
to electric current in winding 1 of the transformer and Figure 4.3b illustrates the magnetic
flux due to electric current in winding 2 of the transformer. The notation in Figure 4.3a
is as follows: ϕ 11 is the total magnetic flux due to current i1. ϕ112 is the magnetic flux
due to current i1 and which links the winding 2. ϕ l1 is the magnetic flux due to electric
current i1 and which “leaks” into air and therefore it does not link winding 2. The
corresponding magnetic fluxes in Figure 4.3b due to current i2 are ϕ 22 , ϕ 221 and ϕ l2 .
Now if the two windings carry currents i1 and i2 simultaneously, the magnetic flux
linkage for windings 1 and 2 can be found by applying superposition (assuming that the
transformer operates in the linear region):
λ1 (t ) = N1ϕ11 (t ) + N1ϕ 221 (t ) = N1ϕ l1 (t ) + N1ϕ112 (t ) + N1ϕ 221 (t )
λ2 (t ) = N 2ϕ 22 (t ) + N 2ϕ112 (t ) = N 2ϕ l 2 (t ) + N 2ϕ112 (t ) + N 2ϕ 221 (t )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
The voltage of the two coils will be:
dϕ (t )
d (ϕ112 (t ) + ϕ 221 (t ))
dλ1 (t )
= r1i1 (t ) + N 1 l1 + N 1
dt
dt
dt
dϕ l 2 (t )
dλ
d (ϕ112 (t ) + ϕ 221 (t ))
v 2 (t ) = r2 i2 (t ) + 2 = r2 i2 (t ) + N 2
+ N2
dt
dt
dt
v1 (t ) = r1i1 (t ) +
Note that the flux leakage ϕ l1 is generated by the electric current i1 and therefore it will
be proportional to this curtrent. Similarly, the flux leakage ϕ l2 is proportional to the
electric current i2, i.e.
ϕ l1 (t ) = a1i1 (t )
ϕ l 2 (t ) = a 2 i2 ( t )
ϕ 11
ϕ 112
i1(t)
i2(t)=0
ϕ l1
v1(t)
v2(t)
(a)
ϕ 221
ϕ 22
i1(t)=0
v1(t)
i2(t)
ϕ l2
v2(t)
(b)
Figure 4.3 Magnetic Flux in a Single Phase Transformer
(a) Primary Coil Energization
(b) Secondary Coil Energization
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By defining
obtain:
φ(t) = ϕ112(t)+ ϕ221(t), and substituting into the above equations, we
di1 (t )
dφ (t )
+ N1
dt
dt
di 2 (t )
dφ (t )
v 2 (t ) = r2 i 2 (t ) + Ll 2
+ N2
dt
dt
v1 (t ) = r1i1 (t ) + Ll1
Where
Ll1 = N1 a1
and
L l2 = N 2 a2
Now let ℜ be the reluctance of the magnetic core. Then
ϕ112 (t ) = N 1i1 (t ) / ℜ
ϕ 221 (t ) = N 2 i 2 (t ) / ℜ
and
ϕ (t ) = ( N 1 i1 (t ) + N 2 i 2 (t )) / ℜ
For magnetic core transformers, the magnetic relunctance ℜ is very small and therefore
N 1i1 (t ) + N 2 i 2 (t ) is also very small. Define an equivalent magnetomotive force in term
of the “magnetizing” current im1:
N1i1 (t ) + N 2 i2 (t ) = N1im1 (t )
Above equation can be rewritten:
N1 (i1 (t ) − im1 (t )) + N 2 i2 (t ) = 0
di (t )
dφ (t ) N 12 di m1 (t )
e1 (t ) = N 1
=
= Lm1 m1
Define
ℜ dt
dt
dt
dφ (t )
e2 (t ) = N 2
dt
2
N
where: Lm1 = 1
ℜ
Upon substitution, the equations of the transformer can be rewritten in the following
form:
di1 (t )
+ e1 (t )
dt
di (t )
v 2 (t ) = r2 i 2 (t ) + Ll 2 2 + e2 (t )
dt
v1 (t ) = r1i1 (t ) + Ll1
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
di m1 (t )
dt
i1 (t ) − im1 (t )
N
=− 2
i2 (t )
N1
e1 (t ) N 1
=
e2 (t ) N 2
e1 (t ) = Lm1
Above equations are represented with the equivalent circuit of Figure 4.4. Note that the
last two equations represent an ideal transformer which is illustrated in Figure 4.4.
i1(t)
r1
Ll 1
i1-im1
Ll 2
r2
i2(t)
im1
v1(t)
Lm1
e1
e2
v2(t)
Ideal Transformer
Figure 4.4 Equivalent Circuit of a Single Phase Transformer
The inductance Lm1 is referred to as the magnetizing inductance and the current im1 as the
magnetizing current. For power transformers under normal operating conditions, the
magnetizing current is very small and many times it is neglected, i.e. the inductance Lm1
is removed from the equivalent circuit. Through a simple transformation, the impedances
can be moved to the other side of the ideal transformer.
Under steady state conditions, the voltages and currents can be expressed in terms of
their phasors (see Chpater 2). In this case, the transformer equations are transformed into:
~
~
~ ~
V1 = r1I1 + jωLl1I1 + E1
~
~
~ ~
V2 = r2 I 2 + jωLl 2 I 2 + E2
~
~
E1 = jωLm1I m1
~ ~
1
I1 − I m1
N
=− 2 =−
~
N1
t
I2
~
E1
N1
=t
~ =
E2 N 2
It is important to note that the resistance and the leakage inductances of a typical
transformer are very small while the magnetizing inductance is very large. Under these
conditions, the leakage impedance of one side can be transferred to the other side. Recall
that by transferring an impedance from one side of the transformer to the other, the
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
impedance is multiplied by the square of the transformation ratio. Because the
magnetizing impedance is very large, the two leakage impedances can be moved to the
same location providing an equivalent leakage impedance. The resulting equivalent
circuit is shown in Figure 4.5. Note that this is an approximation. The error committed by
this approximation is relatively small. Note that the simplified transformer model
equivalent is described with the equivalent series impedance Zeq1, the magnetizing
impedance Zm1 and the transformation ratio t.
~
I1
zeq1
jωLl 1
r1
r2t
2
jωLl 2t
~
V1
2
Ideal
Transformer
~
I2
~
V2
jωLm1
t:1
Figure 4.5 Simplified Equivalent Circuit of a Single Phase Transformer
It is important to note that for the single phase transformer, it is always possible to
develop an equivalent circuit without the presence of the ideal transformer. The
procedure will be illustrated on the simplified equivalent circuit of Figure 4.5. Note that
t=N1/N2. For this purpose, the terminal electric currents are expressed as a function of the
terminal voltages yielding (details are omitted):
~
⎡ I 1 ⎤ ⎡ Yeq1
⎢ ~ ⎥ = ⎢− tY
eq1
⎣I 2 ⎦ ⎣
~
⎤ ⎡V1 ⎤
+ t 2 Ym1 ⎥⎦ ⎢⎣V~2 ⎥⎦
− tYeq1
t 2 Yeq1
where:
Yeq1 =
1
, and
Z eq1
Ym1 =
1
Z m1
The equivalent circuit of Figure 4.6 represents exactly the above equations. The reader is
encouraged to verify that the circuit of Figure 4.6 exhibits the same current/voltage
relationships as the above equations. This can be done by simply expressing the terminal
currents of the circuit of Figure 4.6 as functions of the terminal voltages and comparing
the resulting equations to the above equations.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
~
I1
~
I2
t:1
Zeq=Yeq-1
~
V1
~
V2
Ideal
Transformer
(a)
~
I1
~
V1
~
I2
tYeq
(t2-t)Yeq
(1-t)Yeq
~
V2
(b)
Figure 4.6 Simplified Single Phase Transformer Models
The performance of transformers is characterized with a number of indices. The
following most common indices are introduced:
Transformer efficiency is defined as the ratio of the nominal real power output over the
required real power input, i.e. η = Pout/Pin.
Voltage Regulation is defined as the range of the ratio of the normalized output voltage
magnitude over the normalized input voltage magnitude for loading conditions ranging
from no load to full load. For typical transformers, the ratio of the normalized voltages at
the two ends under no load conditions is approximately 1.0. Therefore the voltage
regulation will be:
R = 1.0 −
V2u
V
V
( full load ) , where: V2u = 2 , and V1u = 1 .
V2 n
V1n
V1u
The subscript n means nominal voltage.
Example E4.2: The equivalent circuit of a 14.4 kV/220 V, 60 Hz, 15 kVA transformer is
illustrated in Figure E4.2. The transformer supplies at the secondary rated power under
rated voltage with power factor 1.0. Compute:
a)
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Compute the voltage at the high side
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
b)
c)
~
I1
~
V1
20
Compute the transformer efficiency, and
What is the transformer voltage regulation
~
I2
~
I
j300
3MΩ
j1.4MΩ
~
E1
~
V1 = 220 e j 0
Rated Load
Power Factor = 10
~
Ib
~
Ig
High Voltage
Side
Low Voltage
Side
Figure E4.2. Parameters of the Transformer of Example E4.2
Solution: The data are marked on Figure E4.2. Note that
o
o
14400
~
E1 =
220e j0 = 14400e j0 volts
220
~ 15000
= 68.1818 A
I2 =
220
o
~
220 ~
I=
I2 = 1.0416e j0 A
14400
a) The voltage at the transformer high side is computed as follows:
o
~ 14, 400 j0o
Ig =
e =.0048e j0 A
3M Ω
o
14
,400 j0 o
~
Ib =
e =.01028e − j90 A
j1.4MΩ
o
o
o
o
~ ~ ~ ~
I1 = I + Ig + Ib = 1.0416e j0 +.0048e j0 +.01028e − j90 = 1.0465e − j.56
o
o
~
V1 = 14, 400 + (20 + j300)(1.0465)e − j.56 = 14, 400 + 314.65e j.85.62
= 14, 427.4e j1.25
o
b) The transformer losses are:
Ploss = (20)(1.0465)2 + (3×106)(.0048)2 = 21.9 + 69.12 = 91.02 Watts
Pout = 15,000 Watts
Pin = 15,091.02 Watts
η = 0.9939
c) The transformer voltage regulation is
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
R = 1.0 −
220 / 220
= 0.00187
14,427 / 14,400
4.3 The Per Unit System
The per unit system is a scaling procedure by which physical quantities are expressed in a
user selected set of units, called the bases. This transformation of physical quantities
results in a transformation of the mathematical models for power apparatus. The basic
idea of the procedure will be described first followed with examples. Then, the per unit
system will be introduced axiomatically.
The basic idea of the per unit system is to express physical quantities in a new system of
units. Consider for example the physical quantity of complex power, S. S is a complex
quantity and the unit of this quantity is VA (volt.ampere). Now assume that we want to
establish a new unit for S which is SB VA, for example SB= 100 VA. Note SB is a scalar.
Now the physical quantity S expressed in the new unit is Su, where Su = S/SB. Note that
Su is now dimensionless. SB is referred to as the base for S and Su is the per unit value of
S. Conceptually, this simple scaling procedure can be applied to all physical quantities.
The bases for all physical quantities can be arbitrarily selected. In practice, however, we
place constraints in the selection of the bases. For example, we require that physical laws
such as Ohm’s law and the power equation are satisfied by the bases, i.e.
VB = Z B I B
S B = VB I B
Where VB, ZB, IB, SB are the bases for voltage, impedance, current and power
respectively. Above requirement limits the number of bases which can be selected
arbitrarily. For the mentioned four quantities (V, Z, I and S), since they must satisfy two
equations (above), then only two quantities can be arbitrarily selected.
Usually, we arbitrarily select SB, VB. Then, the bases IB and ZB are determined from the
power equation and Ohm’s law:
IB =
SB
VB
ZB =
VB VB2
=
I B SB
The above described basic procedure will be illustrated with an example.
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Example E4.3: Consider the simplified circuit of Figure E4.3a. Select as bases for
power and voltage SB = 1000 W and VB = 120 V, respectively. Then express the circuit
equations in the new bases.
r=2Ω
120Vrms
~
I
jx = j2 Ω
Figure E4.3a. A Simple Electric Circuit
Solution:
Since
Then
SB = 1000 W, VB = 120 V
120 2
= 14.4 Ω, and
1000
IB = 8.333 Amps
ZB =
The equation for the circuit of Figure E4.3a is
~ ~
~
V = rI + jxI
Dividing above equation by the voltage base, VB , and recognizing that VB = ZBIB,
~
~
~
jx I
V
r I
=
+
VB Z B I B Z B I B
or
~
~
~
Vu = ru I u + jxu I u
where Vu= 1.0, ru= 0.1389, and xu= 0.1389.
Thus
~
~
1.0 = 0.1389I u + j 0.1389I u
The last equation is represented with the circuit of Figure E4.3b. Note that the circuit of
Figure E4.3b is dimensionless.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
0.1389
1.0
~
I
j0.1389
Figure E4.3b. Scaled Version of the Simple Circuit of Figure E4.3a
4.3.1 Axiomatic Definition of the Per-Unit Scaling System
The per unit system was introduced to simplify the modeling procedure of power
systems. In this respect, there are some desirable characteristics that the per-unit scaling
procedure should possess. These are: (a) The per unit system should be compatible with
other standard systems, such as the metric system. Compatibility means that the physical
laws should be described with the same familiar mathematical expressions, and (b) Bases
should be selected so that scaled values are normally near unity. These requirement are
met with a per unit system which satisfies the following axioms [???]:
Axiom 1: Unscaled equations which describe physical laws shall be identical in form to
their scaled versions. For example, a scaling procedure which converts Ohm's law to the
form "V = kZI," where k ≠ 1 is unacceptable.
Axiom 2: At a given location in a power system, the base value for each quantity to be
scaled (voltage, current, impedance, power, etc.) must be unique.
Axiom 3: Transformer models must be simplified by the scaling procedure.
In the next section, we describe the procedure for selecting a per-unit system that meets
above axioms.
4.3.2 Per-Unit Scaling Procedure
The fundamental scaling equation is:
per − unit value =
actual value
base value
(4.1)
where
actual value
base value
per-unit value
= the unscaled quantity in appropriate SI unit; a phasor or
complex number or a function of time.
= a real number selected to conform to the scaling axioms.
= the scaled dimensionless quantity; a phasor or complex
number, or a function of time.
The term "quantity" here refers to any physical or mathematical entity relating to power
system analysis, including, but not limited to, voltage, current, power, impedance,
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
resistance, inductance, capacitance, charge, flux linkage, time, torque, position (linear
and angular), velocity, and acceleration.
The procedure for selecting the bases for the per-unit system is similar to the process of
selecting the SI system of units. Recall that the SI system of units [???] is based on a
number of arbitrarily selected base units. All other units are expressed as functions of the
base units and are called derived units. Consequently, it is necessary to follow the same
procedure: arbitrarily select a minimum number of bases. These bases shall be called
assigned bases. The remaining bases can be derived and shall be called derived bases.
The derived bases must be selected such as axiom 1 is satisfied. For power engineering
applications, is not necessary to select as assigned bases the full compliment of base SI
units. For all practical power engineering applications, it suffices to select only the
following three assigned bases: (a) power, (b) voltage, and (c) frequency. The derived
bases of interest will be (a) time, (b) current, (c) impedance, (d) inductance, (e)
capacitance, (f) electric charge, and (g) magnetic flux. The procedure is outlined below.
Step 1: Select an arbitrary location in a power system as the reference bus. At this
location, define:
Assigned Bases
fb = frequency base
Vb = voltage base
Sb = power base
= normal operating system frequency, in Hz
= nominal rms line-to-neutral voltage, in volts
= nominal per phase power in VA
Bases for all other physical quantities at this location can be derived from relations
identical to those physical laws which interrelate the unscaled variables.
Derived Bases
tb = time base
=
Ib = current base
=
Zb = impedance base
=
Lb = inductance base
=
Cb = capacitance base
=
Qb = electric charge base
=
Copyright © A. P. Sakis Meliopoulos – 1990-2006
1
fb
Sb
Vb
Vb Vb2
=
I b Sb
Zb
Vb2
=
2πf b S b 2πf b
Sb
1
=
2πf b Z b 2πf bVb2
Sb
C bVb =
2πf bVb
(4.2)
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
λb = magnetic flux base
=
Lb I b =
Vb
2πf b
Step 2: Select the assigned base values at the next location (bus) of the system. Two
cases are recognized:
Case 1: There is no transformer between the present location and another location (for
example bus k) where the bases have already been assigned. In this case the assigned
bases should be the same as the assigned bases at bus k. Note that the derived bases will
be also the same.
Case 2: There is a transformer between the present location and another location (for
example bus k) where the bases have already been assigned. In this case we apply axiom
2 which requires that the transformer model be simplified by the scaling procedure. Let
us examine the transformation of transformer models by the scaling procedure and then
decide how to choose the bases.
Consider a single phase transformer of transformation ratio t = N1/N2. Assume that the
assigned bases have been determined at one side of the transformer, fb1, Sb1, and Vb1. It is
self-understood that the derived bases have been also determined via equations (4.2). We
consider the problem of selecting the assigned bases at the other side (side 2) of the
transformer, i.e., fb2, Sb2, and Vb2. The equations describing the transformer are (for
simplicity the magnetizing current is neglected):
di1 (t )
dφ (t )
+ N1
dt
dt
di 2 (t )
dφ (t )
v 2 (t ) = r2 i 2 (t ) + Ll 2
+ N2
dt
dt
i1 (t )
N2
=−
i2 (t )
N1
v1 (t ) = r1i1 (t ) + Ll1
The per-unitized equations are obtained by dividing the first equation by Vb1, and the
second equation by Vb2. In the process, we can substitute Vb1 with its equal Ib1Zb2 , etc.,
as appropriate yielding:
⎛ i (t ) ⎞
d ⎜⎜ 1 ⎟⎟
v1 (t )
r i (t ) L
⎝ I b1 ⎠ N 1 dφ
= 1 1 + l1
+
Vb1
Z b1 I b1
Lb1 ⎛ t ⎞ Vb1 dt
d ⎜⎜ ⎟⎟
⎝ t b1 ⎠
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(4.3)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
v 2 (t )
r i (t ) L
= 2 2 + l2
Vb 2
Z b2 I b2
Lb 2
⎛ i (t ) ⎞
d ⎜⎜ 2 ⎟⎟
⎝ I b 2 ⎠ N 2 dφ
+
⎛ t ⎞ Vb 2 dt
d ⎜⎜ ⎟⎟
⎝ tb2 ⎠
(4.4)
i1 (t )
I b1
N I
= − 2 b2
i 2 (t )
N 1 I b1
I b2
or
di1u (t ) N 1 dφ (t )
+
dt u
Vb1 dt
di (t ) N dφ (t )
v 2u (t ) = r2u i 2u (t ) + Ll 2u 2u + 2
dt
Vb 2 dt
i1u (t )
N I
= − 2 b2
i 2u (t )
N 1 I b1
v1u (t ) = r1u i1u (t ) + Ll1u
Imposing that the power flow equations will remain invariant (axiom 1) results in the
additional requirements:
Sb2 = Sb1 ,
tb2 = tb1
In this case, Ib1 = Sb1/Vb1, Ib2 = Sb2/Vb2 and therefore Ib2/Ib1 = Vb1/Vb2. Note now that the
i (t)
N / Vb 2
=− 2
last equation becomes iu
. Above equations represent a transformer with
i 2u ( t )
N 1 / Vb1
transformation ratio N1/Vb1 = N2/Vb2 which is illustrated with the circuit of Figure 4.7a.
The circuit of Figure 4.7a is simplified if the bases are selected so that the transformation
ratio of the ideal transformer is 1:1:
N1 N 2
=
Vb1 Vb 2
or
N2
⎛1⎞
Vb1 = ⎜ ⎟Vb1
N1
⎝t ⎠
In this case, per unitized equations of the transformer become:
Vb 2 =
v1u (t ) = r1u i1u (t ) + Ll1u
Copyright © A. P. Sakis Meliopoulos – 1990-2006
(4.5)
di1u (t )
+ eu (t )
dt u
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
v 2u (t ) = r2u i 2u (t ) + Ll 2u
di2u (t )
+ eu (t )
dt
i1u (t )
= −1
i 2u (t )
where:
eu (t ) =
N 1 dφ (t ) N 2 dφ (t )
=
Vb1 dt
Vb 2 dt
Above equations are represented with the circuit of Figure 4.7b. Note that in this case
axiom 3 is satisfied, that is the equivalent circuit of the transformer has been simplified,
the ideal transformer has been removed. However, many times it is not possible to select
the bases according to Equation (4.5). An example will be given later. In this case, the
per unitized transformer model includes an ideal transformer as illustrated in Figure 4.6a.
i1u(t)
r1u
l1u
Ideal
Transformer
l2u
r2u
i2u(t)
v2u(t)
v1u(t)
N1 N2
=
Vb1 Vb2
(a)
i1u(t)
r1u
l1u
l2u
r2u
i2u(t)
v2u(t)
(b)
Figure 4.7. Perunitization of a Transformer
(a) Circuit Realization of Equations (4.3) and (4.4)
(b) Equivalent Circuit in Case N1/Vb1 = N2/Vb2
Above completes the procedures for the selection of the bases.
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
In summary, a practical way to apply above procedure to a power system is as follows:
Classify all power system components as shunt or series components. Shunt
components are those which are connected to one bus only, i.e. generators, loads,
capacitor banks, etc. Series components are those which interconnect two buses.
Examples include power transmission lines, transformers, solid state converters, etc.
Select a bus of the power system, any bus. At this bus (let it be bus i), select the
"assigned bases" as follows:
(1)
(2)
(3)
(4)
The voltage base is selected to be the "nominal" line to ground rms voltage of the
bus.
The power base at bus i is arbitrarily selected to be Sb
The frequency base at bus i is selected to be fb=normal operating system
frequency at bus i.
All other bases (derived) at bus i are computed from Equations (4.2).
Select the assigned bases for all other buses proceeding one bus at a time. Specifically:
(1)
Consider a series component between buses j and k, for which the assigned bases
have been determined at bus j but not at bus k.
(2)
If the series component is: (a) other than transformer: assign bases at bus k same
as at bus j, (b) a transformer with nominal transformation ratio t: assign bases at
this bus as follows:
Sbk = Sbj
Vbk = Vbjt
fbk = fbj
where t is the transformation ratio of the transformer.
(3)
Compute the derived bases at bus k with Equations (4.2).
The per-unitization process is completed when assigned and derived bases have been
defined for all buses in the system.
4.3.3 Mutually Coupled Circuits
Consider two mutually coupled circuits (1 and 2). Assume that the two circuits are
mutually coupled and that the assigned bases have been determined for the two circuits,
i.e. fb1, Sb1, Vb1 and fb2 = fb1, Sb2 = Sb1, and Vb2. This system is illustrated in Figure 4.7.
The equations describing this system are:
~
~
~
~
V1 j − V1k = Z 1 I 1 + Z m I 2
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
~
~
~
~
V2 j − V2 k = Z 2 I 2 + Z m I 1
Above equation can be scaled by dividing the first equation by Vb1 and the second
equation by Vb2, yielding
~
Zm I2
~
~
~
V1 ju − V1ku = Z 1u I 1u +
~
Z b1 I b1
~
Z m I1
~
~
~
V 2 ju − V 2 ku = Z 2u I 2u +
~
Z b2 I b2
~
I1
~
V1 j
~
V2 j
Z1
~
V1k
Zm
~
I2
~
V2 k
Z2
Figure 4.8 Two Mutually Coupled Circuits
In above equations we used the fact that Vb1 = Zb1Ib1 and Vb2 = Zb2Ib2. Now observe that
Z b1 I b1 = I b 2 Z b1 Z b 2
Z b 2 I b 2 = I b1 Z b1 Z b 2
The proof of above relationship can be affected by direct substitution of the equations of
derived base quantities in terms of the assigned base quantities.
Upon substitution in the equation for the coupled circuits:
~
~
~
V1 ju − V1ku = Z 1u I 1u +
~
~
~
V2 ju − V2 ku = Z 2u I 2u +
Zm
Z b1 Z b 2
Zm
Z b1 Z b 2
~
I 2u
~
I 1u
Above equations suggest that the base for the mutual impedance should be
Z bm = Z b1 Z b 2
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Z mu =
Zm
Z bm
Now the equations of the mutually coupled circuits are
~
~
~
~
V1 ju − V1ku = Z1u I1u + Z mu I 2u
~
~
~
~
V2 ju − V2 ku = Z 2u I 2u + Z mu I 1u
The per-unitization process of a system with multiple mutually coupled circuits, such as a
transformer with multiple windings or a synchronous generator, is rather complex. The
per-unitization of a synchronous generator model will be addressed in Chapter 5. The
interested reader should also consult Reference [???] for a comprehensive per-unitization
procedure of the synchronous generator model.
The per unitization procedure will be illustrated with two examples.
Example E4.4: Consider the equivalent circuit of a single phase transformer and a load
as it is illustrated in Figure E4.4. Select a per unit system, and per-unitize the equivalent
circuit of Figure E4.4. The transformer is a 2400V/240V, 5 kVA transformer.
10Ω
j100 Ω
j1Ω
0.1Ω
+
2450 V
11Ω
-6
2x10 Ω
-5
-j10 Ω
j3Ω
-
10:1
Figure E4.4 Equivalent Circuit of a Transformer and Electric Load
Solution: The assigned and derived bases are:
Primary Side:
Vb1 = 2400 V
Sb1 = 5000 VA
5000
= 2.0833 A
Ib1 =
2400
(2400)2
Zb1 =
= 1152 Ω
5000
1
Yb1 =
= .000868055 S
ZB
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Secondary Side:
Vb2 = 240 V
Sb2 = 5000 VA
Ib2 = 20.833 A
Zb2 = 11.52 Ω
Yb2 = .0868055 S
Upon computation of transformer parameters on the new base, the equivalent circuit of
Figure E4.4b is obtained.
j.0868
.00868
.00868
j.0868
+
1.0208
0.95486
.002304
-j.01152
j0.2604
-
Figure E4.4a Per-Unitized Circuit of the System of Figure E4.4
Example E4.5: Consider an example power system. The positive sequence model of this
system is illustrated in Figure E4.5. Assume that the assigned bases at bus 1 are fb1 = 60
Hz, Sb1 = 33.333 MVA, and Vb1 = 7.2 kV.
a) Select the assigned bases at all buses of the system.
b) Compute the per unit positive sequence network of the system.
j0.2 ohms 1 j0.12 ohms
4+j50 ohms
2
3 j13 ohms
4 j0.25 ohms
7.2 kV
10.4 kV
7.2:66.4
66.4:10.4
1.5+j15 ohms
2.5+j28 ohms
5
j12.5 ohms
6
110:115
Figure E4.5 Positive Sequence Network of an Example Power System
Solution: a) The assigned bases for all buses are selected and listed in Table E4.5a.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Table E4.5a Assigned Bases for the Example System of Figure E4.5
Bus
Vb(kV)
Sb(MVA)
fb(Hz)
1
2
3
4
5
6
7.2
66.4
66.4
10.4
66.4
66.4
33.333
33.333
33.333
33.333
33.333
33.333
60
60
60
60
60
60
Note that the derived bases are calculated directly from Equations (4.2). For example, at
bus 6:
Ib = Sb/Vb = 0.50196 kA
Zb = Vb/Ib = 132.282 Ohms
Lb = Zb/2πfb = 0.35089 H
tb = 1/fb = 16.6667 ms, etc.
b) The base impedances are computed using equations (4.2) and listed in Table E4.5b.
Using the bases of Tables E4.5a and E4.5b, the per unit positive sequence network is
computed and shown in Figure E4.5a.
Table E4.5b Derived Impedance Bases for the Example System of Figure
E4.4
Bus
1
2
3
4
5
6
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Zb (ohms)
1.555
132.282
132.282
3.245
132.282
132.282
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
j0.1286
1
j0.077
0.302+j0.378
2
3 j0.098
1.0
4 j0.077
1.0
0.011+j0.1134
0.019+j0.2117
5
j0.0945
6
0.9565:1.0
Figure E4.5a Per Unit Positive Sequence Network of the Example System of
Figure E4.5
4.3.4 The Per Unit System and Numerical Accuracy
The advantages of the per unit system are two: (a) the numerical values of the per unit
voltages indicate how close the voltages are near nominal values, i.e. a per unit value of
0.98 indicates that the voltage is 98% of the nominal value, and (b) the per unit values are
dimensionless. Her we will discuss another less obvious advantage. Specifically, the per
unit system conditions the power system equations in such a way that the numerical
precision of computations on a finite precision computer will be increased. Recall a
recent result [???] which states that the numerical error of a system of equations is
proportional to the matrix condition number k which is defined as the ratio of the largest
to the smallest singular value of the matrix, i.e. k = σmax/σmin. The per unit system results
in system equations with smaller condition number and thus increased numerical
accuracy. This property will be illustrated with an example.
Example E4.6: Consider a 15 kVA, 14.4 kV/220 V, 60 Hz single phase distribution
transformer. The parameters of the transformer are illustrated in Figure E4.6. For
simplicity, the coil resistance will be neglected. The model describing the transformer of
Figure E4.6 (see section ???) is:
di (t )
di1 (t )
+ Lm1 m1
dt
dt
di 2 (t ) Lm1 di m1 (t )
v 2 (t ) = Ll 2
+
dt
t
dt
v1 (t ) = Ll1
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
1
i m1 (t ) = i1 (t ) + i 2 (t )
t
Upon elimination of the magnetizing current, im1(t):
di1 (t )
di (t ) L di (t )
+ Lm1 1 + m1 2
dt
dt
t
dt
di (t ) L di (t ) L di (t )
v 2 (t ) = Ll 2 2 + m1 1 + m2 1 2
dt
t
dt
dt
t
v1 (t ) = Ll1
i1(t)
Ll 1
v1(t)
Ll 2
i2(t)
v2u(t)
Lm1
14400:220
Figure E4.6 Single Phase Distribution Transformer
Assume this model is to be used for transient analysis. Integrating above equations using
the trapezoidal rule, with an integration step of h, we obtain:
⎡ 2( Ll1 + Lm1 )
⎢
h
⎡ v1 (t ) ⎤ ⎢
=
⎢v (t )⎥ ⎢
⎣ 2 ⎦ ⎢
2 Lm1
⎢
th
⎣
2 Lm1
⎤
⎥
th
⎥ ⎧⎡ i (t ) ⎤ ⎡ i (t − h) ⎤ ⎫ ⎡ v1 (t − h) ⎤
Lm1 ⎞ ⎥ ⎨⎢ 1 ⎥ − ⎢ 1
⎛
−
2⎜ Ll 2 + 2 ⎟ ⎣i 2 (t )⎦ ⎣i 2 (t − h)⎥⎦ ⎬ ⎢⎣v 2 (t − h)⎥⎦
⎩
⎭
t ⎠⎥
⎝
⎥
h
⎦
What we would like to do is to study the numerical properties of the matrix before and
after per-unitization. For this purpose, select:
Assigned Bases:
High voltage side:
Power :
15 kVA
Voltage:
14.4 kV
Frequency: 60 Hz
Low voltage side:
Power :
Voltage:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
15 kVA
220 V
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Frequency: 60 Hz
Derived Bases:
High voltage side:
Low voltage side:
Mutual inductance base:
Lb1 = 36.6693 H
Lb2 = 0.00856 H
Lbm= 0.56026 H
Using these bases and an integration time step of h = 100 microseconds, the matrices are:
Actual Units:
⎡74.2802645 1.1347166 ⎤
10 6 × ⎢
⎥
⎣ 1.1247166 0.0173378⎦
Per-Unitized Matrix:
⎡ 33.7613 33.7577 ⎤
10 3 × ⎢
⎥
⎣33.7577 33.7613 ⎦
Upon computation of the singular values of above matrix (using a mathematical
package), we obtain:
Actual Units Matrix:
Largest singular value:
Smallest singular value:
Matrix condition number:
74.2975
0.3694 x 10-5
2.113 x 107
Per-Unitized Matrix:
Largest singular value:
Smallest singular value:
Matrix condition number:
67.519
0.3597 x 10-2
1.8770 x 104
Note that the per-unitization reduced the condition number of the matrix by a factor of
1071.5! (a reduction of three orders of magnitude). This means the numerical truncation
error will be reduced by a factor of 1071.5 using the per-unitized equations [???].
4.3.5 Discussion
The per unit system has three advantages: (1) power system component models are
simplified, for example recall the case of a transformer, (2) the per unit values of voltage
or power may be meaningful, for example a value of 1.0 pu for voltage may mean that
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
the voltage is equal to the nominal voltage, and (3) the model equations are better
conditioned from the numerical point of view. Advantage (2) is very important from the
practical point of view and applies to other quantities as well. Consider for example the
leakage impedance of a transformer. When it is expressed in p.u. (using transformer
ratings as the bases), the pu leakage impedance is typically constant for specific designs.
Specifically, large power transformers have a leakage impedance of 0.07 to 0.1 pu.
Small distribution class transformers have a leakage impedance of about 0.02 to 0.03 pu.,
etc. In other words while the parameters of a transformer may span a large range, when
they are perunitized using the rated values as bases, the parameter values are located in a
relatively narrow range.
4.4 Three Phase Transformers
Three phase transformers can be constructed in a number of ways. Three of the most
usual constructions are illustrated in Figure 4.9. Figure 4.9a illustrates a three phase core
type transformer. The core has three legs, on each leg there are two windings for a total
of six windings. Similarly, Figure 4.9b illustrates a shell type transformer which also has
six windings. Figure 4.9c illustrates a “bank” of three single phase transformers. This
arrangement also has six windings. The six windings of any configuration (a), (b), or (c)
are grouped in two groups of three, the primary and the secondary. For example, in
Figure 4.9a the primary may be the three windings located on the upper part of each leg
and the secondary may be the other three winding. Both the primary and secondary
windings may be connected in a delta or wye configurations leading to four possible
arrangements of a three phase transformer: (a) delta-delta, (b) wye-delta, (c) delta-wye
and (d) wye-wye. These arrangements are schematically represented in Figure 4.10. Note
that from the circuit point of view, all three phase transformer constructions are similar,
i.e. all have six winding grouped into three phases. However the magnetic circuit of each
one of these constructions is different. For example, the three phase transformer bank
consists of three independent magnetic circuits. The shell and core type three phase
transformers are characterized with coupled magnetic circuits of the three phases.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
(a)
(b)
(c)
Figure 4.9 Three Phase Transformers
(a) Core Type
(b) Shell Type
(c) Three Single Phase Transformer Bank
The model of a three phase transformer bank is the simplest since it consists of the
interconnection of three single phase transformers. Replacing each one of the single
phase transformers with its equivalent circuit, the equivalent circuit of the three phase
transformer is obtained. This has been done in Figure 4.11 where the simplified
equivalent circuit of a single phase transformer has been used. The Figure illustrates a
delta-wye connection.
In subsequent paragraphs we will consider first the ideal three phase transformer model
for the purpose of examining its basic characteristics. Then the non-ideal transformer
model will be studied. The use of the symmetrical transformation to the three phase
transformer model will result in the sequence models.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Y
Y
Y Y
Figure 4.10 Schematic Representation of Three Phase Transformers
~
VA
~
VB
~
VC
~
IA
~
I 'a
~
Ia
~
I'b
~
Ib
~
I 'c
~
Ic
~
Va
~
IB
~
Vb
~
IC
~
Vc
α:1
Figure 4.11 Delta-Wye Connected Three Phase Transformer Model
4.4.1 The Ideal Three Phase Transformer
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
An ideal three phase transformer consists of three ideal single phase transformers. The
transformer of Figure 4.10 will be ideal if Y = ∞ (short circuit). The voltage relationships
of an ideal three phase transformer are:
~
~
V AB = aVan
~
~
V BC = aVbn
~
~
VCA = aVcn
Under balanced operating conditions, the voltages will be:
o
~
~
VBn = V An e − j120
o
~
~
VCn = V An e − j 240
o
~
~
Vbn = Van e − j120
o
~
~
Vcn = Van e − j 240
Note that:
o
o
~
~
~
~
~
~
V AB = V An − VBn = V An − V An e − j120 = 3V An e j 30
o
o
o
~
~
~
~
~
~
V BC = V Bn − VCn = V An e − j120 − V An e + j120 = 3V An e − j 90
o
o
o
~
~
~
~
~
~
VCA = VCn − V An = V An e + j120 − V An e − j120 = 3V An e − j 210
Now the relationship between the primary and secondary voltages can be found.
a ~ − j 30o
~
V An =
Van e
3
Æ
a ~ − j 30o
~
V Bn =
Vbn e
3
Æ
a ~ − j 30o
~
VCn =
Vcn e
3
Æ
~
V An
a − j 30 0
e
~ =
V an
3
~
V Bn
a − j 30 0
e
~ =
Vbn
3
~
VCn
a − j 30 0
e
~ =
V cn
3
Above equations indicate that the per phase (positive sequence) equivalent model of a
delta-wye connected ideal three phase transformer is a single phase ideal transformer
with transformation ratio:
~
0
V An
a − j 30 0
n= ~ =
e
= te − j 30
V an
3
Note that in this case the transformation ratio is a complex number.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
4.4.2 Non-Ideal Three Phase Transformer Model
The non-ideal three phase transformer model can be derived from the proper
interconnection of the non-ideal single phase transformers. For simplicity we assume that
each single phase transformer is represented with its simplified non-ideal model. For the
case of a delta-wye connected transformer, the result is illustrated in Figure 4.11.
For the circuit of Figure 4.11, the following relationships are valid:
~
~
~
~
I a' = (V A − V B − aVan )Y
~
~
~
~
I b' = (V B − VC − aVbn )Y
~
~
~
~
I c' = (VC − V A − aVcn )Y
~
~
I a = −aI a'
~
~
I b = −aI b'
~
~
I c = −aI c'
~ ~ ~
I A = I a' − I c'
~ ~ ~
I B = I b' − I a'
~ ~ ~
I C = I c' − I b'
~ ~ ~
~ ~ ~
Upon elimination of the variables E a , E b , E c and Ia' , Ib' , Ic' and expressing the
remaining currents as a function of the voltages we obtain a set of six equations which,
written in matrix notation, are:
~
~
⎡Ia ⎤
⎡a2
0
0 −a a
0 ⎤ ⎡V a ⎤
⎢~ ⎥
⎥⎢ ~ ⎥
⎢
2
0
0 − a a ⎥ ⎢Vb ⎥
⎢Ib ⎥
⎢ 0 a
~
⎢ I~c ⎥
⎢ 0
0 a2
0 − a ⎥ ⎢V c ⎥
a
=
Y
⎢~ ⎥
⎥⎢ ~ ⎥
⎢
2 − 1 − 1 ⎥ ⎢V A ⎥
a
⎢I A ⎥
⎢− a 0
⎢ I~ ⎥
⎢ a − a 0 − 1 2 − 1 ⎥ ⎢V~ ⎥
B
⎢~ ⎥
⎥ ⎢ ~B ⎥
⎢
a − a − 1 − 1 2 ⎦⎥ ⎢⎣VC ⎥⎦
⎢⎣ I C ⎥⎦
⎣⎢ 0
Note that above equation expresses the input/output relationship of the three phase
transformer. In compact matrix form, above equation can be written as
~
⎡ I abc ⎤
⎡ a2I
=
Y
⎢~ ⎥
⎢
T
⎣ − aE
⎣ I ABC ⎦
~
− aE ⎤ ⎡ V abc ⎤
⎥⎢ ~ ⎥
F ⎦ ⎣V ABC ⎦
where I is 3x3 identify matrix, and the matrices E and F are:
⎡ 1 −1 0 ⎤
E = ⎢⎢ 0 1 − 1⎥⎥
⎢⎣− 1 0 1 ⎥⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
⎡ 2 − 1 − 1⎤
F = ⎢⎢− 1 2 − 1⎥⎥
⎢⎣− 1 − 1 2 ⎥⎦
Above equations represent the simplified model of a delta-wye connected three phase
transformer. The same procedure can provide the models for other connections, i.e. deltadelta, wye-wye and wye-delta connections.
4.4.3 Sequence Circuits of Three Phase Transformers
Three phase transformers are inherently symmetric three phase elements. This means
that by applying the symmetrical transformation, their model can be transformed to three
equivalent circuits, namely the positive, negative and zero sequence equivalent circuits.
The procedure will be illustrated on a delta-wye connected transformer model developed
in the previous paragraph. It should be understood that the procedure equally applies to
any other configuration.
The phase voltages and currents are substituted with their corresponding symmetrical
components as follows
~
~
I abc = T −1 I R120
~
~
I ABC = T −1 I L120
~
~
Vabc = T −1V R120
~
~
V ABC = T −1V L120
where R and L indicate right and left side respectively.
Replacing the phase quantities with the symmetrical components, the equation for the
three phase transformer becomes:
~
⎡ I R120 ⎤
⎡ a 2 TIT −1
⎢~ ⎥ = Y ⎢
T
−1
⎣ − aTE T
⎣ I L120 ⎦
~
− aTET −1 ⎤ ⎡V R120 ⎤
⎥⎢ ~ ⎥
TFT −1 ⎦ ⎣V L120 ⎦
Note that by direct evaluation, the following apply:
TIT −1 = I
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
TET −1
⎡ 3e j 30
⎢
=⎢ 0
⎢ 0
⎣
TE T T −1
TFT
−1
o
⎡ 3e − j 30
⎢
=⎢ 0
⎢ 0
⎣
0
o
3e − j 30
0
o
0
o
3e j 30
0
0⎤
⎥
0⎥
0⎥
⎦
0⎤
⎥
0⎥
0⎥
⎦
⎡3 0 0⎤
= ⎢⎢0 3 0⎥⎥
⎢⎣0 0 0⎥⎦
Upon substitution and grouping the six equations into three groups of two we obtain:
o ~
a2
a j 300 ~
~
~
~
3YV R1 −
I R1 = a 2 YV R1 − 3aYe j 30 V L1 =
e 3YV L1
3
3
o ~
a − j 300 ~
~
~
~
I L1 = − 3aYe − j 30 V R1 + 3YV L1 = −
e
3YV R1 + 3YV L1
3
o ~
a2
a − j 300 ~
~
~
~
3YV R 2 −
3YV L 2
I R 2 = a 2 YV R 2 − 3aYe − j 30 V L 2 =
e
3
3
o ~
a j 300 ~
~
~
~
I L 2 = − 3aYe j 30 V R 2 + 3YV L 2 = −
e 3YV R 2 + 3YV L 2
3
a2
~
~
~
I R 0 = a 2 YV R 0 =
3YV R 0
3
~
I L0 = 0
Note that above relations represent three independent set of equations corresponding to
three equivalent circuits which are illustrated in Figure 4.12. The variables appearing in
Figure 4.12 are:
te − j 30 =
0
a
3
e − j 30 , Therefore: t =
0
a
3
The reader is encouraged to verify that the circuits of Figure 4.12 correspond to the above
developed equations. For this purpose, the terminal currents in Figure 4.12 should be
expressed as functions of the terminal voltages and the resulting equations should be
compared to the above equations.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
An example will illustrate the procedure.
~
I L1
~
I R1
3Y
Ideal
Transformer
~
V L1
~
V R1
te − j 30 : 1
0
~
I L2
~
I R2
3Y
Ideal
Transformer
~
VL 2
~
VR 2
0
te j 30 : 1
~
I L0
~
VL 0
~
I R0
3Y
Ideal
Transformer
~
VR 0
t :1
Figure 4.12 Sequence Equivalent of a Delta - Wye Connected Three Phase
Transformer
Example E4.7: A three phase transformer bank is made from three phase transformers.
Each single phase transformer has the equivalent circuit of Figure E4.7a. The three phase
connections are illustrated in Figure E4.7b.
a) Draw the positive sequence equivalent circuit of the three phase transformer bank
with all impedances shown on the left hand side. The transformer ratio and the
impedance values should be clearly marked in actual quantities, i.e. volts and ohms.
b) Draw the positive sequence equivalent circuit of the three phase transformer bank
with all impedances shown on the right hand side. The transformer ratio and the
impedance values should be clearly marked in actual quantities, i.e. volts and ohms.
c) Draw the per phase equivalent circuit in per unit of the three phase transformer bank
using the following bases.
Left Hand Side:
Page 36
Sb = 100 MVA (one phase)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Vb1 = 66.395 kV (line to neutral)
Sb = 100 MVA (one phase)
Vb2 = 6.9282 kV (line to neutral)
Right Hand Side:
j6.6125Ω
j0.024Ω
115kV:
12
kV
3
S=100MVA
(a)
A
a
B
b
C
c
(b)
Figure E4.7 Construction of a Three Phase Transformer Bank
(a) Circuit Model of Single Phase Transformer
(b) Three Phase Connections. Each Block Represents the Single Phase
Transformer of (a)
Solution: The equivalent circuit of Figure E4.7a can be modified by referring the j0.024
ohm leakage impedance on the left hand side. By doing so, the three phase transformer
model becomes identical to the circuit of Figure 4.12 with
Y=
1
S = − j 0.07561 S
j13.225
and
a=
115.0 3
= 16.5988
12.0
a) By utilizing the results of the previous section, the positive sequence equivalent circuit
of the transformer is shown in Figure E4.7a.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
-j0.2268 S
Ideal
Transformer
16 . 5988 e − j 30 : 1
0
Figure E4.7a
b) By referring the admittance on the left hand side, the positive sequence equivalent
circuit of Figure E4.7b is obtained.
-j62.4962 S
Ideal
Transformer
16 . 5988 e − j 30 : 1
0
Figure E4.7b
b) The base admittance at the left hand side is:
S
Yb1 = b21 = 0.02268 S .
Vb1
The per unit value of the admittance is:
Y1u =
− j 0.2268
= − j10.0 .
0.02268
The per unit transformation ratio is:
16.5988
3
e − j 30
0
0
12
: 1 = e − j 30 : 1 .
115
Therefore the per unit equivalent circuit is shown in Figure E4.7c. Similarly the circuit
of Figure E4.7d is developed.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
-j10.0
Ideal
Transformer
e − j 30 : 1
0
Figure E4.7c
-j10.0
Ideal
Transformer
e − j 30 : 1
0
Figure E4.7d
4.5 Autotransformers
Autotransformers are extensively used in power systems for interconnecting two parts of
a system operating at different voltages. Autotransformers are also used in many other
applications. For example, a variac is an autotransformer. Autotranformers can be single
phase devices or three phase devices. Both of these devices will be examined next.
4.5.1 Single-Phase Autotransformers
The construction of a single phase autotransformer is simpler than the usual transformer,
because it requires only one winding. The construction of a single phase autotransformer
is illustrated in Figure 4.13a.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
φ
i1 (t)
+
i2 (t)
A
V1 (t) N1
_
+
N2
V2 (t)
_
(a)
+
V1
+
V2
_
_
(b)
Figure 4.13 Single Phase Autotransformer
(a) Construction
(b) Symbolic Representation
The basic relations of an autotransformer can be derived by assuming that the
autotransformer is ideal. Let ϕ be the magnetic flux inside the core of the
autotransformer of Figure 4.13a. Note that
dϕ (t )
dt
dϕ (t )
v 2 (t ) = N 2
dt
v1 (t ) = N 1
Ρ = ( N 1 − N 2 )i1 (t ) + N 2 (i1 (t ) + i 2 (t )) = ℜϕ (t ) = 0
Forming the ratio of the voltages and currents:
v1 (t ) N 1
=
v 2 (t ) N 2
i1 (t )
N
=− 2
i 2 (t )
N1
The power relationships are
p (t ) = v1 (t )i1 (t ) = − v 2 (t )i 2 (t )
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Above relationships of the ideal autotransformer are also valid for the phasors of the
voltages and currents, i.e.
~
V1
N1
~ =
V2 N 2
~
I1
N2
~ =−
N1
I2
~~
~~
S = V1 I 1* = −V2 I 2*
Note that above equations are identical to the equations of the ideal two winding
transformer.
4.5.2 Three Phase Autotransformers
Three phase autotransformers have magnetic circuits similar to those of the usual
transformers, except that there is only one winding on each magnetic leg of the
transformer with a secondary tap. This winding forms one phase of the three phase
autotransformer. Because of this arrangement and because of the need to ground the
autotransformer, three phase autotransformers are always wye-wye connected. The
analysis of these transformers follows the same procedure as the usual three phase
transformers.
4.6 The Regulating Transformer
The regulating transformer is a transformer arrangement which allows a variable
transformation ratio. This ratio may be a real or a complex number. Typical
constructions of regulating transformers are shown in Figure 4.14. Close examination of
the topology of these transformers will reveal that that transformers 4.14a and 4.14b
simply regulate voltage magnitude while transformer 4.14c regulate both voltage
magnitude and phase. Transformers that regulate both magnitude and phase of the
voltage are referred to as phase shifters.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 41
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
A
a
N
b
C
c
B
(a)
a
A
b
B
c
C
(b)
a
A
b
B
c
C
(c)
Figure 4.14 Example Construction of Regulating Transformers
(a) Variable Tap UnderLoad Transformer
(b) Voltage Magnitude Regulating Transformer
(c) Phase Shifter Transformer
Page 42
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Consider a regulating transformer which regulates both voltage and real power (phase
shift). This transformer is structurally a symmetric three phase element. By modeling
each component of this transformer and subsequent application of the symmetrical
transformation, the positive, negative and zero sequence equivalent circuits can be
developed. The details are omitted. The result is illustrated in Figure 4.15. The
regulating properties of the transformer are expressed with the complex transformation
ratio n = te jα . In Figure 4.15, y is the series admittance of the transformer while ys is the
shunt admittance (magnetizing admittance). It should be clear that the per unit system
can be applied to the models of Figure 4.15 yielding the perunitized positive, negative,
and zero sequence models of the transformer. The models will be identical to those of
Figure 4.15 with the exception that the parameters y, ys, and t will assume their per unit
values. Thus Figure 4.15 illustrates either the actual sequence model or the perunit
sequence model.
~
I L1
~
I R1
y
~
V L1
yS
Ideal
Transformer
~
V R1
te − j 30 : 1
0
~
I L2
~
I R2
y
~
VL 2
yS
Ideal
Transformer
~
VR 2
0
te j 30 : 1
~
I L0
~
I R0
y
~
VL 0
yS
Ideal
Transformer
~
VR 0
t :1
Figure 4.15 Sequence Models of the Phase Shifter
(a) Positive Sequence Model
(b) Negative Sequence Model
(c) Zero sequence Model
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
The voltage/current relationship for the regulating transformer can be developed directly
from the equivalent circuits. As an example consider the positive sequence model (Figure
4.15a). Note that the voltage and current (in per unit) at the right hand side of the ideal
~
~
transformer are V R 2 , and I R 2 . The voltages and currents at the other side of the ideal
transformer will be:
~
te − jα V R 2 and
1 ~
I R 2 respectively.
te jα
These expressions are derived on the basis of zero power consumption inside the ideal
~
~
transformer. Next, and with the help of Figure 4.15a, the currents I L1 and I R1 are
~ ~
expressed in terms of VL1 , V R1 :
~
~
⎡ I L1 ⎤ ⎡ y
− te − jα y ⎤ ⎡VL1 ⎤
⎢~ ⎥ = ⎢
⎥⎢ ~ ⎥
jα
2
⎣ I R1 ⎦ ⎣− te y t ( y + y s )⎦ ⎣VR1 ⎦
(4.6)
For an equivalent circuit to exist, the admittance matrix must be symmetric. Thus only
when α is real, an equivalent circuit exists and it is illustrated in Figure 4.16. The
admittance between node 1 and 2 equals the negative of the entry (1,2) of the admittance
matrix, i.e., the shunt admittance at node 1 will be equal to the sum of entries (1,1) and
(1,2) of the admittance matrix i.e., the shunt admittance at node 2 will be equal to the sum
of the entries (2,1) and (2,2) of the admittance matrix. The modeling of regulating
transformers is illustrated with an example.
~
I L1
~
V L1
ty
(1-t)y
~
I R1
(t2-t)y+t2ys V~R1
Figure 4.16 pi-Equivalent Circuit of the Regulating Transformer of Figure
4.14a, for α=0
Example E4.8: Consider the four bus system of Figure E4.8. The electric load at bus 4 is
2.2 pu. The generating unit 2 generates 1.8 pu real power. The transformer is an off
nominal tap transformer, its reactance is j0.08 pu and the transformer tap is set to 1.05 pu
(bus 4 is the high side). The line data are given in Table E4.8. Derive the positive
sequence equivalent circuit of the system in per unit.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Table E4.8 System Data
Circuit
L1
L2
L3
From Bus
1
2
1
To Bus
2
3
4
Series Admittance
-j10.0
-j9.0
-j12.5
Shunt Admittance
J0.03
J0.02
J0.05
Solution:
The positive sequence pi-equivalent parameters of the off-nominal tap
transformer are :
ys34 = (1.052 - 1.05) (-j12.5) = -j0.65625
ys43 = (1 - 1.05) (-j12.5) = +j0.625
y34 = (1.05) (-j12.5) = -j13.125
The overall positive sequence equivalent circuit is illustrated in Figure E4.8a.
Slack bus
1
G2
2
L1
C1
L2
3
L3
T1
4
SD4
Figure E4.8 A Four Bus Example Power System
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
G2
G1
2
-j10
1
j.02
j.03
j.03
-j9
j.02
j.05
j.02
3
-j12.5
-j0.65625
j.05
-j13.125
j0.625
4
SD4
Figure E4.8a Positive Sequence Equivalent Circuit (in per unit) of the
System of Figure E4.8
Example E4.9: Consider the simplified power system of Figure E4.9. Select bus 1 to be
the reference bus and assume:
S b = 100 MVA / 3, Vb = 15 kV / 3, f b = 60 Hz
All the transmission lines have the same series impedance of 2.0+j23.5 ohms and
negligible shunt impedance. It is desirable to develop the per-unit model of this system.
For this purpose (a) select the assigned bases for the remaining buses 2, 3, 4, 5, and 6, (b)
compute the positive sequence per-unit model of this system.
Page 46
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
250MVA
15kV
60 Hz
1
450MVA
25kV
60 Hz
Y
2
Y
3
4
Z=2.0+j23.5 Ohms
15/230kV
250 MVA
7.5%
230/25kV
450 MVA
8%
Y
6
Y
235/230
150MVA
12%
5
Figure E4.9 A Simplified Power System
Solution: (a) The selected bases appear in Table E4.9. The derived bases can be
computed with the equations that have been presented in section (???).
Table E4.9 Assigned Bases for Simplified Power System of Figure E4.9
Bus
Vb(kV)
Sb(MVA)
fb(Hz)
1
2
3
4
5
6
8.66
132.8
132.8
14.434
132.8
132.8
33.333
33.333
33.333
33.333
33.333
33.333
60
60
60
60
60
60
(b) to be added. (intentionally omitted)
4.7 Transformer Power Flow Equations
In this section we derive the power flow equations for various transformer models
discussed in this chapter. We focus on power flow under steady state balance conditions
and therefore we need to be concerned only with the positive sequence model. In
practice, transformer models of varying accuracy are used based on the desired precision.
Models of varying precision are shown in Figure 4.17.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
y
ysh
(a)
y
te jα : 1
ysh
(b)
2y
2y
te jα : 1
ysh
(c)
2y
1+ t −1
2y
te jα : 1
ysh
(d)
Figure 4.17 Positive Sequence Transformer Model
(a) Non-regulating, (nominal transformation ratio)
(b) Simplified Model of Regulating Transformer
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
(c) Intermediate Model of Regulating Transformer
(d) Realistic Model of Regulating Transformer
The nominal parameters of the transformer (in per unit) are: y s = g + jb , and y sh = jbs .
Note that model (a) is the per unit model of a transformer with fixed tap (nominal tap).
Model (b) is a simplified model of an off-nominal tap transformer. The simplification
consists of assuming that the leakage impedance of the transformer is independent of the
tap setting and that it is all on the left side of the transformer. Model (c) is a simplified
model of an off-nominal tap transformer. The simplification consists of assuming that
the leakage impedance of the transformer is independent of the tap setting. Model (d) is
the most accurate transformer model. The leakage impedance on the non-tapped side is
constant while the leakage impedance on the tapped side is proportional to the quantity
(1+abs(t-1)) where t is the tap setting. This model results from considering the
construction characteristics of the variable tap transformers. For accurate power flow
computations, model (d) should be used.
The power flow equations for the four models are:
Power flow equations for model 1.
THESE EQUATIONS ARE WRONG- Update
P1 = gV12 − V1V2 [ g cos(δ 1 − δ 2 ) + b sin(δ 1 − δ 2 )]
Q1 = −(b + bs )V12 + V1V2 [b cos(δ 1 − δ 2 ) − g sin(δ 1 − δ 2 )]
P2 = gV22 − V1V2 [ g cos(δ 2 − δ 1 ) + b sin(δ 2 − δ 1 )]
Q2 = −bVk22 + V1V2 [b cos(δ 2 − δ 1 ) − g sin(δ 2 − δ 1 )]
Power flow equations for Model 2.
P1 = t 2 gV12 − V1V2 t[ g cos(δ 1 − δ 2 + α ) + b sin(δ 1 − δ 2 + α )]
Q1 = −(t 2b + bS )V12 + V1V2t[b cos(δ 1 − δ 2 + α ) − g sin(δ 1 − δ 2 + α )]
P2 = gV22 − V1V2 t[ g cos(δ 2 − δ 1 − α ) + b sin(δ 2 − δ 1 − α )]
Q2 = −bV22 + V1V2 t[b cos(δ 2 − δ 1 − α ) − g sin(δ 2 − δ 1 − α )]
Power flow equations for Model 3.
P1 = (2 gtA + (2bt + bs ) B)V12 − 2tV1V2α12 A + 2tV1V2 β12 B
Q1 = (2 gtB − (2bt + bs ) A)V12 − 2tV1V2α12 B − 2tV1V2 β12 A
P2 = (2 gA + (2b + bs ) B)V22 − 2tV1V2α 21 A + 2tV1V2 β 21B
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Q2 = (2 gB − (2b + bs ) A)V22 − 2tV1V2α 21B − 2tV1V2 β 21 A
where:
α12 = g cos(δ 1 − δ 2 + α ) + b sin(δ 1 − δ 2 + α )
β12 = g sin(δ 1 − δ 2 + α ) − b cos(δ 1 − δ 2 + α )
α 21 = g cos(δ 2 − δ 1 − α ) + b sin(δ 2 − δ 1 − α )
β 21 = g sin(δ 2 − δ 1 − α ) − b cos(δ 2 − δ 1 − α )
A=
4( g 2 + b 2 )(1 + t 2 ) + 2bbs
4( g 2 + b 2 )(1 + t 2 ) 2 + bs2 + 4bbs (1 + t 2 )
B=
2 gbs
4( g + b )(1 + t ) + bs2 + 4bbs (1 + t 2 )
2
2
2 2
Power flow equations for Model 4.
P1 = (2 gtA + (2bt + sbs ) B)V12 − 2tV1V2α12 A + 2tV1V2 β12 B
Q1 = (2 gtB − (2bt + sbs ) A)V12 − 2tV1V2α12 B − 2tV1V2 β12 A
P2 = (2 gA + (2b + bs ) B)V22 − 2tV1V2α 21 A + 2tV1V2 β 21B
Q2 = (2 gB − (2b + bs ) A)V22 − 2tV1V2α 21B − 2tV1V2 β 21 A
where:
α12 = g cos(δ 1 − δ 2 + α ) + b sin(δ 1 − δ 2 + α )
β12 = g sin(δ 1 − δ 2 + α ) − b cos(δ 1 − δ 2 + α )
α 21 = g cos(δ 2 − δ 1 − α ) + b sin(δ 2 − δ 1 − α )
β 21 = g sin(δ 2 − δ 1 − α ) − b cos(δ 2 − δ 1 − α )
A=
Page 50
4( g 2 + b 2 )( s + t 2 ) + 2 sbbs
4( g 2 + b 2 )( s + t 2 ) 2 + s 2bs2 + 4 sbbs ( s + t 2 )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
B=
2sgbs
4( g + b )( s + t ) + s 2bs2 + 4sbbs ( s + t 2 )
2
2
2 2
⎧ t , if t ≥ 1.0
s=⎨
⎩2 − t , if t ≤ 1.0
4.8 Transformer Parameter Identification From Tests
Actual modeling of transformers requires that their parameters be accurately determined.
For this purpose, tests can be performed to measure the parameters of the transformer.
Two simple tests are quite usual for this purpose: (a) the open circuit test and (b) the
short circuit test.
4.8.1 The Open Circuit Test
The open circuit test is shown in Figure 4.18. Specifically Figure 4.18a shows
connections and measurements for an open circuit test of a single phase transformer.
Note that the secondary of the transformer is open, while the primary is connected to a
source. The voltage at the terminals of the transformer is typically near the nominal
value. The voltage at the primary side is measured (V), the electric current of the primary
is also measured (A) as well as the real power flow into the transformer (W).
Transformer
Under Test
W
A
Open
V
(a)
Transformer
Under Test
r1
W
jxl1
A
V
g
r2
jxl2
-jb
Open
(b)
Figure 4.18 Open Circuit Test
(a) Setup for the Open Circuit test
(b) Equivalent Circuit of the Open Circuit Test
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Consider the measurements of voltage, current and power, Voc, Ioc and Poc respectively.
By replacing the transformer under test with its equivalent circuit, as it is shown in Figure
4.18b, the following relationship can be derived:
P
g ≅ oc2
V oc
⎛I
b = ⎜⎜ oc
⎝ Voc
I
g − jb ≅ oc
Voc
2
⎞
Poc2
⎟⎟ − 4
⎠ Voc
Therefore this test provides the magnetizing reactance and the core loss of the
transformer.
4.8.2 The Short Circuit Test
The short circuit test is shown in Figure 4.19. Specifically, Figure 4.19a shows
connections and measurements for a short circuit test of a single phase transformer. Note
that the secondary of the transformer is shorted, while the primary is connected to a
source. The applied voltage at the terminals of the transformer is adjusted so that the
current flowing into the transformer is typically near the nominal value. The voltage at
the primary side is measured (V), the electric current of the primary is also measured (A)
as well as the real power flow into the transformer (W).
Transformer
Under Test
Rated
amps
W
A
V
(a)
Transformer
Under Test
jxl1
r1
W
A
V
g
r2
jxl2
-jb
(b)
Figure 4.19 Short Circuit Test
(a) Setup for the Short Circuit test
(b) Equivalent Circuit of the Short Circuit test
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Consider the measurements of voltage, current and power, Vsc, Isc and Psc respectively.
By replacing the transformer under test with its equivalent circuit as it is shown in Figure
4.19b, the following relationships can be derived:
r1 + a 2 r2 ≅
Psc
I sc2
(
)
r1 + a 2 r2 + x l1 + a 2 x l 2 ≅
xl1 + a x l 2
2
⎛V
= ⎜⎜ sc
⎝ I sc
Vsc
I sc
2
⎞
P2
⎟ − sc4
⎟
I sc
⎠
Therefore this test provides the leakage impedance and equivalent resistance of the
transformer. Note that this test does not provide the impedances of the individual coils
(high and low side) but rather the overall equivalent.
4.9 Normalized Power System Model
Delta-wye connected three phase transformers are extensively used in three phase power
systems for a variety reasons. These transformers generate a 30o phase shift of the
voltage phase angles. Systems are designed in such a way that these phase shifts are
consistent in the sense that for any loop in the network the net phase shift will be zero. In
this case, we call the system normal. In terms of modeling the system, the 30o phase
shift of the delta-wye connected transformers can be ignored if the system is normal.
However, if it is necessary to compute the actual system voltages, these phase shifts must
be modeled.
Consider for example the system of Figure 4.20. The Figure illustrates a typical system.
Note that for this system, there is no network loop in which the net phase shift by deltawye (or wye-delta) connected transformers is different than zero. There are two deltawye connected transformers that are connected radially. Thus the system is normal. In
modeling this system, the 30o phase shift of the delta-wye connected transformers are
omitted for simplicity.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
18kV:230kV
230kV
230kV:115kV
∆Y
YY
∆Y
Auto
25kV:230kV
115kV:13.8kV
∆Y
230kV:115kV
∆Y
115kV:13.8kV
Figure 4.20 Example Normal Power System
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
4.10 Summary and Discussion
In this chapter, the modeling of power transformers has been addressed. The power
transformer under normal operating conditions behaves as an approximately linear
device. Three phase transformers are symmetric and therefore can be represented with
their positive, negative and zero sequence networks. We developed these models for
three phase transformers. These models are useful for power flow analysis, short circuit
analysis and similar analysis problems. Regulating transformers were also addressed and
their models were discussed.
In this chapter we also introduced the per unit system. The per unit system is useful for
simplifying transformer models, expressing power system quantities in a meaningful way
and conditioning the power system model for better numerical stability.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
4.11 Problems
Problem P4.1: A single phase transformer has the following ratings: 10-kVA, 60 Hz,
2300V/230V. The impedances of the transformer are:
r1 = 4.2 ohms ,
r2 = 0.042 ohms
xl1 = 5.5 ohms,
x l 2 = 0.055 ohms,
Lm1 = 17.0 Henries (on 2300V side)
Subscript 1 denotes the 2300 V winding, subscript 2 the 230 V winding. The transformer
delivers 90% of its rated volt-amperes (apparent power) at a 0.85 power factor (current
lagging) to a load on the low voltage side with 230 V across the load. The core losses are
70 Watts.
•
•
•
•
Compute the total leakage impedance referred to (1) the high voltage side, and (2)
the low voltage side.
Compute the high side terminal voltage,
Compute the power factor at the high side terminals,
Compute the transformer efficiency.
Solution: 1st step:
Compute ZL
(230 V) IL = (.90)(10,000) VA
~
VL = 230e j.0
⇒
IL = 39.13 A
cos(ϕV - ϕI) = .85
ϕ V − ϕ I =.5548
⇒
~
IL = 39.13e − j.5548
~
VL
Z L = ~ = 5.8778 Ω e j.5548 = R L + jXL = 4.996 Ω + j3.0963 Ω
IL
~
V1 = [ ( 499 + (2)( 4.2)) + j((2)(5.5) + 309.63)] 3.913e − j.5548
= (507.4 + j 320.63)(3.913) e − j.5548
= 2348.6423 V e − j.00875
⇒
b)
V1 = 2,348.6423 V
Power factor
pf = cos(φ V1 , I ' L ) = cos(.00875 + .5548) ⇒
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
pf = .84536
c)
Efficiency = 1 -
Losses
Pout
Pout = (230)(39.13)(.85) = 7649.9 Watts
Losses = 70 W + (2)(4.2)(3.913)2 = 198.62 Watts
Efficiency = .974
Problem P4.2: Consider the three winding transformer of Figure P4.2. An electric
current i1(t)= 210A cos ωt , ω = 377 sec-1 is injected into winding 1. Winding 2 is loaded
with a series L, C circuit while winding 3 is open, i.e., i3(t) = 0. The values of L and C
are: L=0.5H, C=10µF. The number of turns are: N1=20, N2=40, N3=100. Assume ideal
magnetic material ( µ = ∞ ) and ideal windings (zero resistance). Compute the voltage
v1(t) and v3(t) across the windings 1 and 3.
i2
i1
L
N
+
2
V1
N
1
C
+
N3
i 3 V3
Figure P4.2
Solution:
N 1i1 ( t ) + N 2 i 2 ( t ) + N 3 ⋅ 0 = 0
N1
i1 (t ) = − 2 5 A cos ωt
N2
1 ~
~
~
V 2 = − jωLI 2 −
I 2 = − j 383.76 V
jωC
⇒ i 2 (t ) = −
Recall
Copyright © A. P. Sakis Meliopoulos – 1990-2006
~
~
V2 = jωN 2 Φ
~
Φ = −0.02545T
Φ(t ) = 2 0.02545T cos(ωt − 180o )
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
V1 (t ) = N 1
dΦ
= 2191.89V cos(ωt − 90 o )
dt
V3 (t ) = N 3
dΦ
= 2 959.46V cos(ωt − 90 o )
dt
Problem P4.3: A single phase distribution transformer has the following parameters:
rated voltages 14.4 kV/220 V, rated power 37.5 kVA, rated frequency 60 Hz, leakage
impedance 3.7%, total winding resistance 0.5% .
(a)
(b)
(c)
Compute and draw the equivalent circuit of the transformer including an ideal
transformer. Refer all impedances to the high voltage side.
Compute and draw the pi equivalent circuit of the transformer in actual units.
Compute and draw the per unit equivalent circuit of the transformer assuming
the following bases on the high voltage side of the transformer: Vb = 14.4 kV,
Sb = 37.5 kVA. Select the low side bases as you like.
Problem P4.4: Given a 125 MVA, 230kV/115kV three phase, wye/wye connected
transformer with 10% leakage impedance and negligible resistance. The transformer has
several taps on the 115 kV side. Assume that the tap of the transformer is set for 110 kV.
Compute the per unit positive sequence equivalent circuit of the transformer. Use as
bases the following (nominal voltages): High voltage side: Vb = 230 / 3 kV, Sb = 125/3
MVA, low voltage side: Vb = 115 / 3 kV, Sb = 125/3 MVA.
Solution: The per phase equivalent circuit in pernuit is shown in Figure 1. Elimination of
the ideal transformer yields the circuit of Figure 2.
Problem P4.5: A 230 kV transmission line and a 115 kV line parallel each other for a
distance of 12 miles. The zero sequence equivalent model of the system is illustrated in
Figure P4.5. The assigned bases for the 230 kV line are:
f b, 230 = 60 Hz, S b, 230 = 100 MVA / 3, Vb, 230 = 230 kV / 3
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The assigned bases for the 115 kV line are:
f b,115 = 60 Hz, S b,115 = 100 MVA / 3, Vb,115 = 115 kV / 3
Compute the per unit zero sequence equivalent model of this system. What is the base
impedance for the mutual impedance?
1 ohms
j210 ohms
230 kV Line
j10 ohms
115 kV Line
j28 ohms
1.5 ohms
Figure P4.5
Problem P4.6: Consider the electric power system of Figure P4.6. The transformer
reactances are given in per unit on each transformer ratings. The generator controls the
voltage at bus 1 to nominal value. Assume that the indicated electric load is a constant
impedance load and it absorbs the indicated amount of power when the voltage is
nominal. Compute and draw the positive sequence equivalent circuit of the system in per
unit using a 100/3 MVA power basis (one phase) and nominal voltages.
G1
1
3
2
Y
Z=6+j62Ω
45MVA
15kV/115kV
X=8%
4
Y
30MVA
115kV/25kV
X=7.5%
Sd 4 =20MW+
j10MVAR
Figure P4.6
Solution: The base impedance at the three different kV levels of this system are:
2
(
15)
=
= 2.25 ohms
100
(115)2 = 132.25 ohms
Z b,115kV =
100
(25)2 = 6.25 ohms
Z b, 25kV =
100
The transformer impedances referred to 100 MVA basis are:
Z b,15kV
z T 1 = j 0.08
100
= j 0.1778
45
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
z T 1 = j0.075
100
= j 0.25
30
The positive sequence equivalent circuit is:
j0.178
0.0454
j0.469
j0.25
4
1.0pu
j2
1:e
j300
1:e
-j300
Problem P4.7. Consider the phase-shifter transformer of Figure P4.7. Note that the phase
shifter transformer is made of six identical single-phase transformers. Assume that each
transformer is ideal and the primary to secondary transformation ratio is 0.5.
What is the positive sequence equivalent of this phase-shifter transformer?
What is the negative sequence equivalent of this phase-shifter transformer?
What is the zero sequence equivalent of this phase-shifter transformer?
Hint: Apply the appropriate voltage (positive, negative, or zero sequence) at one side of
the phase-shifter transformer and compute the voltages at the other side.
S
A
a
P
P P
S
S
P
B
C
S
P
S
S
b
c
P
Phase Shifter
Figure P4.7
Problem P4.8: An electric power system consists of a transmission line and a
transformer as it is illustrated in Figure P4.8. Compute the admittance matrix of the
system comprising the line and the transformer, i.e. neglect the presence of the generator
and electric load. All relevant values are given in the Figure.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
G1
3
2
1
Z=j0.1 pu
1:t
Z=j0.12 pu
t =1.05
Figure P4.8
Problem P4.9. Figure P4.9 illustrates a simplified power system consisting of two
synchronous machines, a phase shifting transformer and a line. Assume that the
synchronous machines are ideal voltage sources and that the phase shifter consists of a
number of ideal transformers. Specifically, all illustrated transformers are ideal with a
transformation ratio of secondary to primary voltage equal to 0.5 (Secondary
Voltage/Primary voltage = 0.5) and they are connected as illustrated in the figure. Note
that the primary is designated as P and the secondary as S. (a) Compute the phase to
ground voltage (both magnitude and phase) at the two sides of the phase shifter. (b)
Compute the total electric power absorbed or generated by the synchronuous machine 2.
The following additional information is given.
0
0
0
~
~
~
~
~
E1a = 8.66e j 0 kV , E1b = E1a e − j120 , E1c = E1a e − j 240
0
0
0
~
~
~
~
~
E 2 a = 9.0e j 0 kV , E 2b = E 2 a e − j120 , E 2 c = E 2 a e − j 240
S
~
E 1a
~
E 1c
y = 0.2 - j1.0 mhos
~
E 2a
P
~
E 1b
P P
S
S
P
y = 0.2 - j1.0 mhos
~
E 2b
~
E 2c
S
P
Synchronuous
Machine 1
S
S
y = 0.2 - j1.0 mhos
P
Phase Shifter
Transmission
Line
Synchronuous
Machine 2
Figure P4.9
Solution:
The voltage at phase A at left hand side of the line is:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
(
)
0
~ ~
~
~
E = E1a − 0.25 E1b − E1c = 9.437e j 23.41 kV
The transformation ratio is:
~
0
E
n = ~ = 1.0897 e j 23.41
E1a
The electric current of the line (going into the synchronous machine 2) is:
(
)
0
~
~ ~
I = (0.2 − j1.0) E − E 2 a = 3.839e j16.49 kA
The power is:
~ ~
S 2 = (3)E 2 a I * = 99.41 MW − j 29.42 MVAr
Synchronous machine absorbs real power therefore operates as motor.
Problem P4.10. Consider the normalized model of the off nominal tap transformer of
Figure P4.10. This tap setting of the ransformer corresponds to a normalized
transformation ratio of t=1.09 pu as it is indicated in the figure. All indicated admittance
values are in pu. Under the present operating conditions, the voltages at the terminals of
the transformer are (magnitude in pu, phase in radians):
~
~
V1 = 1.0e − j 0.15 , V2 = 0.98e − j 0.25
Compute the complex power flow at the left side of the transformer.
-j10.9
-j10
+
+
~
~
V2
V1
1.09 : 1
Figure P4.10
Problem P4.10: The per phase positive sequence equivalent circuit of a three phase
transformer is illustrated in Figure P4.10. At a certain instant of time the voltages at the
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
terminals of the transformer are as illustrated in the figure. Compute the total (threepahse) power P1 + j Q1 at the left hand side of the transformer.
Ω
-j9.18
Ω
-j0.1
P1+jQ1
o
o
63e j0 kV
7.2 e j18 kV
o
11512
: e j30 kV
Figure P4.10
Solution:
~
Assume voltage at high side of ideal transformer to be E . Then
(
)
0
12 j 300 ~ ⎞
~
⎛
I 2 = − j 9.18⎜ 7.2 j18 −
e E⎟
115
⎝
⎠
~
~
I 1 = − j0.1 63 − E ,
115 j 300 ~
~
I2 = −
e I1
12
~
Upon substitution of the currents we obtain one equation in E . Solution of this equation
yields:
0
~
E = 65.2458 − j 7.1714 = 65.6387e − j 6.27 kV
Then:
(
)
0
~
~
I 1 = − j 0.1 63 − E = 0.71714 + j 0.22458 kA = 0.7515e j17.39 kA , and
0
115 j 300 ~
~
I2 = −
e I 1 = 7.2019e − j132.61 kA
12
The power at the two ends is:
(
S 1 = P1 + jQ1 = 3(63.0) 0.7515e j17.39
(
)(
0
)
*
= 142.0335e − j17.39 MVA = 135.415MW − j 42.4503MVAr
0
S 2 = P2 + jQ 2 = 3 7.2e j18 7.2019e − j132.61
0
0
)
*
= 155.5611e j150.61 MVA = −135.415MW + j 76.3419MVAr
0
Note: the difference in real powers uis due to numerical precision errors.
Problem P4.11: Consider the two models of an off-nominal tap transformer of Figure
P4.11, the model of Figure P4.11a is the approximate model while the model of Figure
P4.11b is an intermediate accuracy model.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
(a) Assume that the applied voltages at the two terminals of the transformer are:
0
0
~
~
V1 = 1.025e j 0 , V 2 = 1.005e − j 5 . Compute the complex power flow from side 1 to side 2
for the two transformer models. Assume that the model shown in Figure P4.11b is the
accurate model and compute the percent error of the real and reactive power flow for the
model of Figure P4.11a.
(b) Compute the pi-equivalent circuits of the two transformer models in Figures P4.11a
and P4.11b.
j0.10
+
~
V1
+
~
V2
1:1.05
(a)
j0.05
+
~
V1
j0.05
+
~
V2
1:1.05
(b)
Figure P4.11
Solution: (a) The complex power for model a is:
(
(
~
~ ~
S 12 a = 1.05V1 − j10.0 1.05V1 − V 2
))
*
= 1.12246e j 40.38 = 0.85505 + j 0.72719
0
The complex power for model b is:
~~
S 12b = V1 I 1*
Where the current will come from
(
)
~
~ ~
I 1 = − j 20.0 V1 − E , and
~
I1
~ ~
= − j 20.0 1.05E − V 2
1.05
(
)
Upon solution for the current on side 1:
(
~
~ ~
I 1 = − j9.988109 1.05V1 − V2
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Therefore:
(
(
~~
~
~ ~
S 12b = V1 I 1* = V1 − j 9.988109 1.05V1 − V 2
))
*
= 1.1484745e j 40.38 = 0.87486 + j 0.74404
0
The error is:
error =
S 12 a − S 12 b
= −2.3%
S 12b
(b) The pi-equivalent of model a is:
-j10.5
-j0.525
j0.5
The pi-equivalent of model b is:
-j9.9881
-j0.499405
j0.475624
Problem P4.12: Consider the three phase electric power system of Figure P4.12
consisting of two ideal synchronous machines G1 and G2, an ideal delta-wye connected
transformer and a transmission line. The indicated generated voltages are:
o
~
E1 = 7.2 kV e -j15 ,
o
~
E 2 = 14.9 kV e j0
Compute the total real and reactive power delivered or absorbed by the synchronous
machine G1. Repeat for the synchronous machine G2. Which machine operates as a
generator and which as a motor?
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Unit G1
~
E1
S1 = P1 + jQ 1
2+j10Ω
+
_
2+j10Ω
Unit G2
+
_ ~
E
2
2+j10Ω
1:1
Figure P4.12
Solution:
o
~
V1 = 7.2 3 kV e j15
o
o
~
I 1 = 7.2 3 e j15 − 14.9 3 e j0 (2 + j10)
~
I 1 = (− 2.8542 + j3.2277) (2 + j10)
= 0.32277 + j0.28542
(
)
= 0.4309 kA e − j41.5
o
Transformer is ideal. Thus
~
S1 = 3V1 I1*
= (3)(7.2)( 3 )(0.4309) e − j26.5 MVA
o
= 16.121 e − j26.5 MVA
= 14.427 MW -j7.193 MVAR
o
Unit 1 is operating as a generator.
Problem P4.13: The following data apply to a single phase 1000 kVA 66kV/6.6kV, 60
Hz transformer:
Frequency
Voltage
Current
Power
with low-voltage
terminals short-circuited
60 Hz
3240 V
(high voltage
side)
15.2 A
(high
voltage side)
7,490 W
With high-voltage
terminals open
60 Hz
6600 V
(low voltage
side)
9.1 A
(low voltage
side)
9,300 W
Assume that the primary and secondary resistances are equal when referred to the same
side and that the primary and secondary leakage reactances are similarly equal.
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
a)
b)
Determine the equivalent circuit of the above transformer with all parameter values
referred to the high side,
Determine the equivalent circuit of the above transformer with all parameter values
expressed in a per unit system. Use the ratings of the transformer as bases.
Solution:
1)
Short Circuit Test
(r1 + r2 ) + ( jX1 + X2 ) =
3240
= 213.15 Ω
15.2
(r1 + r2)(15.2)2 = 7490 ⇒ r1 + r2 = 32.418 Ω
X1 + X2 = 213.15 2 − 32.418 2 = 210.67 Ω
referred to H.S.
r1 = r2 = 16.21 Ω
X1 = X2 = 105.34 Ω
Open Circuit Test
rf =
6600 2
= 4683.87 Ω
9300
1
6600
=
g − jb
9.1
⇒
⇒ g = .0002134 S
2
⎛ 9.1 ⎞
2
b= ⎜
⎟ − g =.001362 S
⎝ 6600 ⎠
Therefore
2)
ZB =
VB VB2 (66000) 2
=
=
= 4356 Ω
IB
S B 1000000
r2' pu = r1pu =
X'2 pu = X1pu
YB =
16.21
=.00372
4356
105.34
=
=.02418
4356
1
ZB
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
g pu =
.000002134
=.00929
YB
b pu =.05932
Problem P4.14. Consider the simplified three-phase power system illustrated in Figure
P4.14. The parameters of each component are given below.
At a certain instant of time, the overall system operates under balanced conditions and
the induction motor operates with a slip of 0.04. Compute:
a) the electric current on Phase A of the induction motor terminal.
b) the total real power absorbed by the induction motor.
Wye-Wye Connected Transformer
1
SOURCE
G
2
HIGHSIDE
LOWSIDE
Three-Phase Cable
Three-Phase Source
IM
SPEED
Induction Motor
Figure P4.14
SOURCE:
Voltage: 4.16 kV, Line to Line
Assume ideal voltage source
CABLE:
Per phase (positive sequence) impedance: 0.69+j0.69 ohms
TRANSFORMER
Rated Transformation Ratio: 4.16kV/480 Volts
Power Rating: 500 kVA
Equivalent Impedance: j1.38 ohms referred to the high side
Core Loss, Magnetizing current, winding losses: neglect
INDUCTION MOTOR
Power Rating: 500 kVA
Voltage rating: 480 Volts, Line to Line
Per Phase Stator resistance: 0.0046 ohms
Per Phase Rotor resistance: 0.0092 ohms referred to stator
Per Phase Stator leakage reactance: 0.0138 ohms
Per Phase Rotor leakage reactance: 0.0184 ohms referred to stator
Page 68
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Core loss, magnetizing reactance: neglect
Problem P4.15. Consider the model of the “phase-shifter” transformer of Figure P4.15.
All indicated transformers are identical. Note that the taps can move to different positions
– all transformers at the same tap position by virtue of the mechanical link that is
indicated by the dashed line. Note also that the centers of the secondary windings are
bonded (solid line). Assume that at the indicated position (of the tap) the transformation
ratio of the transformers is 0.06, i.e.
V s , AB V s , AC
=
= ... = 0.06
V AB
V AC
Assume that all transformers are ideal and compute:
a) the positive sequence transformation ratio
b) the negative sequence transformation ratio, and
c) the zero sequence transformation ratio.
~
~
VS,AB
VS,AC
A
a
1~
V
2 AB
1 ~
V
2 AC
B
b
1~
V
2 BC
1~
V
2 BA
c
C
1~
V
2 CA
1~
2 VCB
Figure P4.15
Solution: Given the current setting of the tap:
~
~
~
~
~
~
~
V a = V s , AC − V s , AB + V A = V A + 0.03V B − 0.03VC
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
Positive sequence:
sequence. Then:
Assume voltages on left hand side are balanced and positive
~
Va
− j 2.9745 0
~ = 1.00134 e
VA
~
VA
O
~
0.03VB
~
-0.03VC
~
Va
Negative sequence:
sequence. Then:
Assume voltages on left hand side are balanced and negative
~
Va
j 2.9745 0
~ = 1.00134 e
VA
Zero sequence: Assume voltages on left hand side are balanced and zero sequence. Then:
~
Va
~ = 1. 0
VA
Problem P4.16: Consider the model of the off nominal tap transformer of Figure P4.16.
The indicated transformation ratio t is in p.u. and y is the transformer leakage admittance
in p.u. Note that the leakage admittance of the tapped coil is inveresely proportional to
the tap setting t. Compute the equivalent circuit of this transformer. The equivalent
circuit should contain only admittance elements, i.e. the ideal transformer should be
eliminated.
y/t
y
t:1
Figure P4.16
Solution:
Page 70
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
The equations relating the terminal voltages and currents are (see designations on
Figure):
Figure
~ y ~ y ~
I 1 = V1 − Et
t
t
Problem P4.17. Consider the normalized model of the off nominal tap transformer of
Figure P2. This tap setting of the ransformer corresponds to a normalized transformation
ratio of t=1.09 pu as it is indicated in the figure. All indicated admittance values are in
pu. Under the present operating conditions, the voltages at the terminals of the
transformer are (magnitude in pu, phase in radians):
~
~
V1 = 1.0e − j 0.67 , V2 = 0.98e − j 0.25 , phase angles are in radians
Compute the complex power flow at the left side of the transformer.
-j10.9
~
V1
j0.05
-j10
Ideal
Transformer
te
− j 300
~
V2
:1
Figure P2
Solution: Let E be the voltage on the right of the ideal transformer. Then:
(
0 ~
~
~
I 1 = − j10.9 V1 − 1.09e − j 30 E
(
~
~ ~
I 2 = − j10.0 V2 − E
)
)
(
)
~ ~
j10.0 V 2 − E
~
− j 30 0 ~
− j 30 0 ~
j10.9 V1 − 1.09e
E + j (0.05)(1.09 )e
E+
=0
0
1.09e j 30
(
)
Solution of last equation for E:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 4, Meliopoulos
0 ~
0 ~
~
E = 0.5444606e j 30 V1 + 0.458261e j 30 V 2
Substitution in the equations for the terminal currents yields:
(
)
(
)
~
~
~
I 1 = − j 4.431263V1 − 5.4445989V2
~
~
~
I 2 = − j − 5.444606V1 + 5.41739V 2
The power at the left side of the transformer is:
~~
S 12 = V1 I 1* = −2.17569 − j0.440712 pu
Page 72
Copyright © A. P. Sakis Meliopoulos – 1990-2006