Transcript Lecture 10

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
[email protected]
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
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
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
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
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
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
Per Unit Example, cont’d
Z BLeft
8kV 2

 0.64
100 MVA
Z BMiddle
Z BRight
80kV 2

 64
100 MVA
16kV 2

 2.56
100 MVA
Same circuit, with
values expressed
in per unit.
7
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*
8
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
Actual
VL
 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
9
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!
10
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
11
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
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!
13
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
14
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
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
OriginalBase
Z pu

2
Vbase
/
OriginalBase
S Base
NewBase
S Base
OriginalBase
S Base

2
Vbase
NewBase
S Base
NewBase
 Z pu
NewBase
Z pu
16
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
17
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
18
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
3f Transformer Interconnections
D-Y
Y-D
20
Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
VAn
VAB
IA 1
 a,
 a,

Van
Vab
Ia a
21
Y-Y Connection: 3f Detailed Model
22
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
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
24
D-D Connection: 3f Detailed Model
To use the per phase equivalent we need to use
the delta-wye load transformation
25
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
D-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
27
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
28
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
Y-D Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a
phase shift of -30 degrees.
30
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
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
32
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
33
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
34
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