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1. ECE 476
Power System Analysis
Lecture 10: Per Unit, Three-Phase
Transformers, Load and Generators
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu

2. Announcements
• Please read Chapter 2.4, Chapter 6 up to 6.6
• HW 4 is due now; HW 5 is 5.31, 5.43, 3.4, 3.10, 3.14,
3.19, 3.23, 3.60, 6.30 should be done before exam 1
• Exam 1 is Thursday Oct 6 in class
• Closed book, closed notes, but you may bring one 8.5 by 11
inch note sheet and standard calculators
• Last name A-M here, N to Z in ECEB 1013
• Power Area Scholarships, Awards (Oct 1 deadlines,
except Nov 1 for Grainger)
• http://energy.ece.illinois.edu/
• Turn both into Prof Sauer (4046 or 4060)
1

3. Service Entrance Grounding
Image: www.osha.gov/dte/library/electrical/electrical_10.gif
We’ll
talk more
about
grounding
later in
the semester
when we
cover faults
2

4. 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

3

5. Per Unit Conversion Procedure, 1f
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)
4

6. 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
5

7. 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
6

8. Per Unit Example, cont’d
2
2
2
8
0.64
100
80
64
100
16
2.56
100
Left
B
Middle
B
Right
B
kV
Z
MVA
kV
Z
MVA
kV
Z
MVA
  
  
   Same circuit, with
values expressed
in per unit.
7

9. 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
 
    

       
   
  
       
8

10. Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
L
Actual
Actual
L
Actual
G
Middle
B
Actual
Middle
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 MVA
I 1250 Amps
80 kV
I 0.22 30.8 Amps 275 30.8
V
S
S
       
     
     
 
       
9

11. Three Phase Per Unit
1. Pick a 3f VA base for the entire system,
2. Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
3. Calculate the impedance base
Procedure is very similar to 1f except we use a 3f
VA base, and use line to line voltage bases
3
B
S f
2 2 2
, , ,
3 1 1
( 3 )
3
B LL B LN B LN
B
B B B
V V V
Z
S S S
f f f
  
Exactly the same impedance bases as with single phase!
10

12. Three Phase Per Unit, cont'd
4. Calculate the current base, IB
5. Convert actual values to per unit
3 1 1
3 1
B B
, , ,
3
I I
3 3 3
B B B
B LL B LN B LN
S S S
V V V
f f f
f f
   
Exactly the same current bases as with single phase!
11

13. 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!
12

14. 3f 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
 
    

       
   
  
       
Again, analysis is exactly the same!
13

15. 3f Per Unit Example, cont'd
L
Actual
Actual
L
Actual
G
Middle
B
Actual
Middle
0.859 30.8 27.6 kV 23.8 30.8 kV
0.189 0 300 MVA 56.7 0 MVA
0.22 30.8 300 MVA 66.0 30.8 MVA
300 MVA
I 125 (same cur
0 Amps
3138 kV
I 0.22 30.
rent!)
8
V
S
S
       
     
     
 
    Amps 275 30.8
    
Differences appear when we convert back to actual values
14

16. 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
15

17. 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
2 2
Hence Z /
Z
base base
OriginalBase NewBase
pu actual pu
OriginalBase NewBase
pu pu
OriginalBase NewBase
Base
Base
NewBase
OriginalBase NewBase
Base
pu pu
OriginalBase
Base
Z Z Z
V V
Z
S
S
S
Z
S
 
 
 
16

18. Per Unit Change of Base Example
•A 54 MVA transformer has a leakage reactance of
3.69%. What is the reactance on a 100 MVA base?
100
0.0369 0.0683 p.u.
54
e
X   
17

19. 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)?
2
100
0.10 0.0286 p.u.
350
230
0.0286 15.1
100
e
X   
  
18

20. Three Phase Transformers
• There are 4 different ways to connect 3f
transformers
Y-Y D-D
Usually 3f transformers are constructed so all windings
share a common core
19

21. 3f Transformer Interconnections
D-Y Y-D
20

22. Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
1
, ,
An AB A
an ab a
V V I
a a
V V I a
  
21

23. Y-Y Connection: 3f Detailed Model
22

24. 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.
23

25. D-D Connection
Magnetic coupling with AB/ab, BC/bb & CA/ca
1 1
, ,
AB AB A
ab ab a
V I I
a
V I a I a
  
24

26. D-D Connection: 3f Detailed Model
To use the per phase equivalent we need to use
the delta-wye load transformation
25

27. 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.
26

28. D-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
27

29. D-Y Connection V/I Relationships
, 3 30
30
30
Hence 3 and 3
For current we get
1
1
3 30 30
3
1
30
3
AB AB
an ab an
an
An
AB
ab an
AB
a AB
ab
A AB AB A
a A
V V
a V V V
V a
V
V
V V
a a
I
I a I
I a
I I I I
a I
     
 
 
 
  
       
   
28

30. 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
29

31. Y-D Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a
phase shift of -30 degrees.
30

32. 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"
31

33. LTCs and Circulating Vars
slack
1 1.00 pu
2 3
40.2 MW
40.0 MW
1.7 Mvar
-0.0 Mvar
1.000 tap 1.056 tap
24.1 MW
12.8 Mvar
24.0 MW
-12.0 Mvar
A
MVA
1.05 pu
0.98 pu
24 MW
12 Mvar
64 MW
14 Mvar
40 MW
0 Mvar
0.0 Mvar
80%
A
MVA
32

34. Phase Shifting Transformers
• Phase shifting transformers are used to control the
phase angle across the transformer
– Also called phase angle regulators (PARs) or quadrature
booster transformers
• 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.
33
Image Source: en.wikipedia.org/wiki/Quadrature_booster#/media/File:Qb-3ph.svg

35. Phase Shifter Example 3.13
slack
Phase Shifting Transformer
345.00 kV 341.87 kV
0.0 deg
216.3 MW 216.3 MW
283.9 MW 283.9 MW
1.05000 tap
39.0 Mvar 6.2 Mvar
93.8 Mvar
125.0 Mvar
500 MW
164 Mvar 500 MW
100 Mvar
34

36. 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
auto-transformers 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!
35