ECE 476
Power System Analysis
Lecture 10: Per Unit, Transformers,
Load, Generators
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Please read Chapter 3; start on Chapter 6
• H5 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 2.38, 6.9
•
It should be done before the first exam, but does not need
to be turned in
• First exam is Tuesday Oct 6 during class
•
Closed book, closed notes, but you may bring one 8.5 by
11 inch note sheet and standard calculators.
1
Service Entrance Grounding
We’ll
talk more
about
grounding
later in
the semester
when we
cover faults
Image: www.osha.gov/dte/library/electrical/electrical_10.gif
2
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. Pick a 3f VA base for the entire system, S B3f
2. Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
3. 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!
3
Three Phase Per Unit, cont'd
4. Calculate the current base, IB
I3Bf
S B3f
3 S 1Bf
S 1Bf



 I1Bf
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
4
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!
5
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!
6
3f Per Unit Example, cont'd
Differences appear when we convert back to actual values
V LActual  0.859  30.8  27.6 kV  23.8  30.8 kV
S LActual  0.1890  300 MVA  56.70 MVA
SGActual  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
7
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
8
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
9
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
X e  0.0369 
 0.0683 p.u.
54
10
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
11
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
12
3f Transformer Interconnections
D-Y
Y-D
13
Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
VAn
VAB
IA 1
 a,
 a,

Van
Vab
Ia a
14
Y-Y Connection: 3f Detailed Model
15
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.
16
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
17
D-D Connection: 3f Detailed Model
To use the per phase equivalent we need to use
the delta-wye load transformation
18
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.
19
D-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
20
D-Y Connection V/I Relationships
VAB
VAB
 a,
 Van  Vab  3 Van30
Van
a
VAn30
VAB 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
 a  a I A30
3
1

I A30
3
21
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
22
Y-D Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a
phase shift of -30 degrees.
23
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"
24
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
25
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.
Image Source: en.wikipedia.org/wiki/Quadrature_booster#/media/File:Qb-3ph.svg
26
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
27
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!
28
Load Models
• Ultimate goal is to supply loads with electricity at
constant frequency and voltage
• Electrical characteristics of individual loads matter,
but usually they can only be estimated
–
–
actual loads are constantly changing, consisting of a large
number of individual devices
only limited network observability of load characteristics
• Aggregate models are typically used for analysis
• Two common models
–
–
constant power: Si = Pi + jQi
constant impedance: Si = |V|2 / Zi
29
Generator Models
• Engineering models depend upon application
• Generators are usually synchronous machines
• For generators we will use two different models:
–
–
a steady-state model, treating the generator as a constant
power source operating at a fixed voltage; this model
will be used for power flow and economic analysis
a short term model treating the generator as a constant
voltage source behind a possibly time-varying reactance
30
Power Flow Analysis
• We now have the necessary models to start to
develop the power system analysis tools
• The most common power system analysis tool is the
power flow (also known sometimes as the load
flow)
–
–
–
–
power flow determines how the power flows in a network
also used to determine all bus voltages and all currents
because of constant power models, power flow is a
nonlinear analysis technique
power flow is a steady-state analysis tool
31
Linear versus Nonlinear Systems
A function H is linear if
H(a1m1 + a2m2) = a1H(m1) + a2H(m2)
That is
1)
the output is proportional to the input
2)
the principle of superposition holds
Linear Example: y = H(x) = c x
y = c(x1+x2) = cx1 + c x2
Nonlinear Example: y = H(x) = c x2
y = c(x1+x2)2 ≠ (cx1)2 + (c x2)2
32