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8/18/2019 lec2 power system
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POWER SYSTEMS ILecture 2
06-88-590-68
Electrical and Computer Engineering
University of Windsor
Dr. Ali Tahmasebi
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1
Per Phase Analysis
l Per phase analysis allows analysis of balanced 3f systems with the same effort as for a single phase
system
l Balanced 3f Theorem: For a balanced 3f systemwith
– All loads and sources Y connected
– No mutual Inductance between phases
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2
Per Phase Analysis, cont’d
l Then
– All neutrals are at the same potential
– All phases are COMPLETELY decoupled
– All system values are the same sequence as sources. Thesequence order we’ve been using (phase b lags phase a
and phase c lags phase a) is known as “positive”
sequence; later in the course we’ll discuss negative and
zero sequence systems.
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3
Per Phase Analysis Procedure
To do per phase analysis
1. Convert all D load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)5. If necessary, go back to original circuit to determine
line-line values or internal Dvalues.
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Per Phase Example
Assume a 3f , Y-connected generator with Van = 1Ð 0°volts supplies a D-connected load with ZD = -jWthrough a transmission line with impedance of j0.1W
per phase. The load is also connected to aD-connected generator with Va”b” = 1Ð 0°through asecond transmission line which also has an impedanceof j0.1W per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each generator, SY and SD
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Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
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Per Phase Example, cont’d
' ' 'a a a
To solve the circuit, write the KCL equation at a'1
(V 1 0)( 10 ) V (3 ) (V j3
j j- Ð - + + - Ð - 30°)(- 10 )=0
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Per Phase Example, cont’d
' ' 'a a a
'a
' 'a b
' 'c ab
To solve the circuit, write the KCL equation at a'
1(V 1 0)( 10 ) V (3 ) (V j
3
10(10 60 ) V (10 3 10 )3
V 0.9 volts V 0.9 volts
V 0.9 volts V 1.56
j j
j j j j
- Ð - + + - Ð - 30°)(- 10 ) =0
+ Ð °= - +
= Ð - 10.9° = Ð - 130.9°
= Ð 109.1° = Ð 19. volts1°
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8
Per Phase Example, cont’d
*'*
ygen
*
" '"
S 3 5.1 3.5 VA0.1
3 5.1 4.7 VA0.1
a aa a a
a agen a
V V V I V j
j
V V S V j j
D
æ ö-= = = +ç ÷
è ø
æ ö-= =- -ç ÷è ø
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9
Power in Balanced 3 Circuits
In this balanced three-
phase circuit, T can be a
voltage source, a motor
or any impedance load.
For phase ‘a’: = 2 cos +
= 2 cos +
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10
Power in Balanced 3 Circuits, cont’d
d - b = q is the phase shift between phase voltage and
phase current.
= () .()
= cos − + cos 2 + +
∅ = () + () + () = 3 cos − =
∅ (NOT a function of time)
In terms of line-to-line voltages:
∅ = 3 cos − = ∅
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Power in Balanced 3 Circuits, cont’d
• For a generator, delivered power is calculated from
these 2 equations, for a load, these equations
calculate absorbed power.
• Advantage: for example, power delivered to a 3Fac motor is always constant, which results in
constant speed and torque. In contrast, a single
phase ac motor may have torque pulsations.
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12
Power in Balanced 3 Circuits, cont’d
Complex power:
For phase ‘a’:
= Ð d , ̅ = Ð b ® ̅ = ̅∗ = Ð (d-b)
̅∅ = ̅ + ̅ + ̅ = 3 ̅= 3 ̅∗ = 3Ð (d−
b)
= 3 cos + sin = 3 cos + sin
̅∅
= ∅
+ ∅
Total apparent power:
̅∅ = ∅ = 3= 3 [VA]
NOTE: ̅ is a complex number but not a phasor!
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Power System Components
l All power systems have three major components:
Generation, Load and Transmission/Distribution.
l Generation: Creates electric power.
l Load: Consumes electric power.
l Transmission/Distribution: Transmits electric power
from generation to load.
– Lines/transformers operating at voltages above 100 kV
are usually called the transmission system. The
transmission system is usually networked.
– Lines/transformers operating at voltages below 100 kV
are usually called the distribution system (radial).
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Power Transformers
l Power systems are characterized by many different
voltage levels, ranging from 765 kV down to
240/120 volts.
l Transformers are used to convert voltage and transfer power between different voltage levels.
l The ability to inexpensively change voltage levels is
a key advantage of ac systems over dc systems.
l In this section we’ll development models for the
transformer and discuss various ways of connecting
three phase transformers.
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Ideal Transformer
l First we review the voltage/current relationships for
an ideal transformer
– no winding resistance ® no real power losses
– magnetic core has infinite permeability
– no leakage flux
– no core loss
l We’ll define the “primary” side of the transformer
as the side that usually takes power, and thesecondary as the side that usually delivers power.
– primary is usually the side with the higher voltage, but
may be the low voltage side on a generator step-up
transformer.
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Ideal Transformer Equations
1 1 2 2
1 21 1 2 2
1 2 1 1
1 2 2 2
Assume we have flux in magnetic material. Then
= turns ratio
m
m m
m m
m
N N
d d d d v N v N dt dt dt dt
d v v v N a
dt N N v N
f
l f l f
l f l f
f
= =
= = = =
= = ® = =
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Current Equations in Ideal Transformer
'1 1 2 2
'
1 1 2 2'
1 1 2 2
'1 1 2 2
To get the current relationships use ampere's law
mmf
length
length
Assuming uniform flux density in the core
lengtharea
d N i N i
H N i N i
B N i N i
N i N i
m
f
m
G= = +
´ = +
´= +
´ = +´
òH L
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Current/Voltage Equations
'1 1 2 2
1 2 1 2'
1 2 12
1 2
1 2
If is infinite then 0 . Hence
1or
Then
0
10
N i N i
i N i N
N i N ai
av v
i i
a
m = +
= - = =
é ùé ù é ùê ú=ê ú ê úê úë û ë û
û
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19
Impedance Transformation Example
• Example: Calculate the primary voltage and current
for an impedance load on the secondary
21
21
01
0
a vv
vi Z a
é ùé ùé ù ê úê ú=ê ú ê úë û ê úûû
21 2 1
21
1
1 vv a v ia Z
va Z
i
= =
=
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Real Transformers
• Real transformers
– have losses
– have leakage flux
– have finite permeability of magnetic core
• Add R1
and R2
to windings 1 and 2 to represent i2 R
losses of each winding
• Add X 1 to account for leakage flux of winding 1
(flux that links winding 1 but not winding 2)
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Transformer Core Losses
• Eddy currents arise because of changing flux in core.
Eddy currents are reduced by laminating the core
• Hysteresis losses are proportional to area of BH curve
and the frequency
These losses are reduced
by using material with a
thin BH curve
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Transformer Core Losses
• To represent both eddy current loss and Hysteresis
loss, we add a resistive shunt branch R C (or GC) that
carries the current iC = core loss current.
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Effect of Finite Core Permeabili ty
m
1 1 2 2 m
m 21 2
1 1
2 m1 2 m1 1
Finite core permeability means a non-zero mmf
is required to maintain in the core
N
This value is usually modeled as a magnetizing current
where im
i N i
N i i
N N
N i i i N N
f
f
f
f
- = Â
Â= +
Â= + =
Add X m
to represent this effect.
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Transformer Equivalent Circuit
Using the previous relationships, we can derive an
equivalent circuit model for the real transformer
' 2 '2 2 1 2
' 2 '2 2 1 2
This model is further simplified by referring all
impedances to the primary side
r e
e
a r r r r
x a x x x x
= = +
= = +
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Simplified Equivalent Circuit
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Calculation of Model Parameters
• The parameters of the model are determined based
upon
– nameplate data: gives the rated voltages and power
– open circuit test: rated voltage is applied to secondarywith primary open; measure the secondary current and
losses (Wattage).
– short circuit test: with secondary shorted, apply voltage
to primary to get rated current to flow at primary;
measure voltage and losses.
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Transformer Example
Example: A single phase, 100 MVA, 200/80 kV
transformer has the following test data:
open circuit: 20 amps, with 10 kW losses
short circuit: 30 kV, with 500 kW losses
Determine the model parameters.
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Transformer Example, cont’d
e
2
sc e
2 2e
2
e
100 30500 , R 60
200 500
P 500 kW R 2 ,
Hence X 60 2 60
200410
200R 10,000 10,000
20
sc e
e sc
c
e m m
MVA kV I A jX
kV A
R I
kV R M kW
kV jX jX X
A
= = + = = W
= = ® = W= - = W
= = W
+ + = = W =
From the short circuit test
From the open circuit test
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Residential Distribution Transformers
Single phase transformers are commonly used in
residential distribution systems. Most distribution
systems are 4 wire, with a multi-grounded, common
neutral.
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Per Unit Calculations
l A key problem in analyzing power systems is the
large number of transformers.
– It would be very difficult to continually have to refer
impedances to the different sides of the transformersl This problem is avoided by a normalization of all
variables.
l This normalization is known as per unit analysis.
actual quantityquantity in per unit
base value of quantity=
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Per Unit Conversion Procedure, 1
1. Pick a 1f VA base for the entire system, SB
2. Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.3. Calculate the impedance base, ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unit
Note, per unit conversion on affects magnitudes, not
the angles. Also, per unit quantities no longer have
units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
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Per Unit Solution Procedure
1. Convert to per unit (p.u.) (many problems are
already in per unit)
2. Solve
3. Convert back to actual as necessary
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Per Unit Example
Solve for the current, load voltage and load power
in the circuit shown below using per unit analysis
with an SB of 100 MVA, and voltage bases of
8 kV, 80 kV and 16 kV.
Original Circuit
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Per Unit Example, cont’d
2
2
2
80.64
100
8064
100
162.56
100
Left B
Middle B
Right B
kV Z
MVA
kV Z
MVA
kV Z
VA
= = W
= = W
= = W
Same circuit, with
values expressed
in per unit.
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Per Unit Example, cont’d
L
2*
1.0 0 0.22 30.8 p.u. (not amps)3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
L L L L
G
I j
V S V I
Z
S
Ð °= = Ð - °+
= Ð °- Ð - °́ 2.327Ð 90°
= 0.859Ð - 30.8°
= = =
= Ð °́ Ð °=0.22Ð °
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Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
L
Actual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 16 kV 13.7 30.8 kV
0.189 0 100 MVA 18.9 0 MVA
0.22 30.8 100 MVA 22.0 30.8 MVA
100 MVAI 1250 Amps80 kV
I 0.22 30.8 Amps 275 30.8
V
S
S
= Ð - °́ = Ð - °= Ð °́ = Ð °
= Ð °́ = Ð °
= =
= Ð - °́ 1250 = Ð - °A