Transcript ECE 310 - Course Website Directory

ECE 476
POWER SYSTEM ANALYSIS
Lecture 9
Transformers, Per Unit
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements


Be reading Chapter 3
HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class.
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Transformer Equivalent Circuit
Using the previous relationships, we can derive an
equivalent circuit model for the real transformer
This model is further simplified by referring all
impedances to the primary side
r2'  a 2 r2
re  r1  r2'
x2'  a 2 x2
xe  x1  x2'
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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 primary with
secondary open; measure the primary current and losses
(the test may also be done applying the voltage to the
secondary, calculating the values, then referring the
values back to the primary side).
short circuit test: with secondary shorted, apply voltage
to primary to get rated current to flow; measure voltage
and losses.
4
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
From the short circuit test
100 MVA
30 kV
I sc 
 500 A, R e  jX e 
 60 
200kV
500 A
2
Psc  Re I sc
 500 kW  R e  2 ,
Hence X e  602  22  60 
From the open circuit test
200 kV 2
Rc 
 4M 
10 kW
R e  jX e  jX m
200 kV

 10, 000 
20 A
X m  10, 000 
6
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

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 transformers
This problem is avoided by a normalization of all
variables.
This normalization is known as per unit analysis.
actual quantity
quantity in per unit 
base value of quantity
8
Per Unit Conversion Procedure, 1f
1.
2.
3.
4.
5.
Pick a 1f VA base for the entire system, SB
Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.
Calculate the impedance base, ZB= (VB)2/SB
Calculate the current base, IB = VB/ZB
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.
2.
3.
Convert to per unit (p.u.) (many problems are
already in per unit)
Solve
Convert back to actual as necessary
10
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
11
Per Unit Example, cont’d
Z BLeft
8kV 2

 0.64
100 MVA
Middle
ZB
Z BRight
80kV 2

 64
100 MVA
16kV 2

 2.56
100 MVA
Same circuit, with
values expressed
in per unit.
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Per Unit Example, cont’d
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8p.u.
VL I L*
13
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
V LActual  0.859  30.8 16 kV  13.7  30.8 kV
S LActual  0.1890 100 MVA  18.90 MVA
SGActual  0.2230.8 100 MVA  22.030.8 MVA
I Middle

B
100 MVA
 1250 Amps
80 kV
I Actual
Middle  0.22  30.8 Amps  275  30.8
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Three Phase Per Unit
Procedure is very similar to 1f except we use a 3f
VA base, and use line to line voltage bases
1.
2.
3.
Pick a 3f VA base for the entire system, S B3f
Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
Calculate the impedance base
ZB 
VB2, LL
S B3f

( 3 VB, LN ) 2
3S 1Bf

VB2, LN
S 1Bf
Exactly the same impedance bases as with single phase!
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Three Phase Per Unit, cont'd
4.
Calculate the current base, IB
3f
IB
S B3f
3 S 1Bf
S 1Bf
1f



 IB
3 VB, LL
3 3 VB , LN VB , LN
Exactly the same current bases as with single phase!
5.
Convert actual values to per unit
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Three Phase Per Unit Example
Solve for the current, load voltage and load power
in the previous circuit, assuming a 3f power base of
300 MVA, and line to line voltage bases of 13.8 kV,
138 kV and 27.6 kV (square root of 3 larger than the
1f example voltages). Also assume the generator is
Y-connected so its line to line voltage is 13.8 kV.
Convert to per unit
as before. Note the
system is exactly the
same!
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3f Per Unit Example, cont'd
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8p.u.
*
VL I L
Again, analysis is exactly the same!
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3f Per Unit Example, cont'd
Differences appear when we convert back to actual values
VL
 0.859  30.8 27.6 kV  23.8  30.8 kV
Actual
SL
SGActual
 0.1890 300 MVA  56.70 MVA
Actual
 0.2230.8 300 MVA  66.030.8 MVA
I Middle

B
300 MVA
 1250 Amps (same current!)
3 138 kV
I Actual
Middle  0.22  30.8  Amps  275  30.8
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3f Per Unit Example 2
Assume a 3f load of 100+j50 MVA with VLL of 69 kV
is connected to a source through the below network:
What is the supply current and complex power?
Answer: I=467 amps, S = 103.3 + j76.0 MVA
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Per Unit Change of MVA Base


Parameters for equipment are often given using
power rating of equipment as the MVA base
To analyze a system all per unit data must be on a
common power base
NewBase
Z OriginalBase

Z

Z
pu
actual
pu
Hence ZOriginalBase

pu
ZOriginalBase

pu
2
Vbase
/
OriginalBase
S Base
NewBase
S Base
OriginalBase
S Base
2
Vbase
NewBase
S Base
NewBase
 Z pu
NewBase
 Z pu
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Per Unit Change of Base Example
A 54 MVA transformer has a leakage reactance or
3.69%. What is the reactance on a 100 MVA base?
100
X e  0.0369 
 0.0683 p.u.
54
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Transformer Reactance


Transformer reactance is often specified as a
percentage, say 10%. This is a per unit value
(divide by 100) on the power base of the
transformer.
Example: A 350 MVA, 230/20 kV transformer has
leakage reactance of 10%. What is p.u. value on
100 MVA base? What is value in ohms (230 kV)?
100
X e  0.10 
 0.0286 p.u.
350
2
230
0.0286 
 15.1 
100
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Three Phase Transformers
There are 4 different ways to connect 3f transformers
D-D
Y-Y
Usually 3f transformers are constructed so all windings
share a common core
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3f Transformer Interconnections
D-Y
Y-D
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Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
VAn
VAB
IA 1
 a,
 a,

Van
Vab
Ia a
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Y-Y Connection: 3f Detailed Model
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Y-Y Connection: Per Phase Model
Per phase analysis of Y-Y connections is exactly the
same as analysis of a single phase transformer.
Y-Y connections are common in transmission systems.
Key advantages are the ability to ground each side
and there is no phase shift is introduced.
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D-D Connection
Magnetic coupling with AB/ab, BC/bb & CA/ca
VAB
I AB 1 I A 1
 a,
 ,

Vab
I ab a I a a
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D-D Connection: 3f Detailed Model
To use the per phase equivalent we need to use
the delta-wye load transformation
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D-D Connection: Per Phase Model
Per phase analysis similar to Y-Y except impedances
are decreased by a factor of 3.
Key disadvantage is D-D connections can not be
grounded; not commonly used.
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D-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
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D-Y Connection V/I Relationships
VAB
VAB
 a,
 Van  Vab  3 Van30
Van
a
VAB 30
VAn30
Hence Vab  3
and Van  3
a
a
For current we get
I AB 1
  I a  a I AB
I ab a
I A  3 I AB   30  I AB
1

I A30
3
1
 a  a I A30
3
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D-Y Connection: Per Phase Model
Note: Connection introduces a 30 degree phase shift!
Common for transmission/distribution step-down since
there is a neutral on the low voltage side.
Even if a = 1 there is a sqrt(3) step-up ratio
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Y-D Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a
phase shift of -30 degrees.
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Load Tap Changing Transformers




LTC transformers have tap ratios that can be varied
to regulate bus voltages
The typical range of variation is 10% from the
nominal values, usually in 33 discrete steps
(0.0625% per step).
Because tap changing is a mechanical process, LTC
transformers usually have a 30 second deadband to
avoid repeated changes.
Unbalanced tap positions can cause "circulating
vars"
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LTCs and Circulating Vars
slack
64 MW
14 Mvar
1.00 pu
1
24.1 MW
12.8 Mvar
40.2 MW
1.7 Mvar
1.000 tap
A
A
80%
1.056 tap
MVA
MVA
40.0 MW
-0.0 Mvar
24.0 MW
-12.0 Mvar
0.98 pu
2
3
1.05 pu
0.0 Mvar
24 MW
12 Mvar
40 MW
0 Mvar
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Phase Shifting Transformers



Phase shifting transformers are used to control the
phase angle across the transformer
Since power flow through the transformer depends
upon phase angle, this allows the transformer to
regulate the power flow through the transformer
Phase shifters can be used to prevent inadvertent
"loop flow" and to prevent line overloads.
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Phase Shifter Example 3.13
345.00 kV
500 MW
341.87 kV
283.9 MW
39.0 Mvar
283.9 MW
6.2 Mvar
slack
164 Mvar
Phase Shifting Transformer
216.3 MW
125.0 Mvar
500 MW
100 Mvar
216.3 MW
0.0 deg
93.8 Mvar
1.05000 tap
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ComED Control Center
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ComED Phase Shifter Display
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Autotransformers



Autotransformers are transformers in which the
primary and secondary windings are coupled
magnetically and electrically.
This results in lower cost, and smaller size and
weight.
The key disadvantage is loss of electrical isolation
between the voltage levels. Hence autotransformers are not used when a is large. For
example in stepping down 7160/240 V we do not
ever want 7160 on the low side!
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