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 [email protected]
Jan 13, 2016
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
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.
2
Service Entrance Grounding
Image: www.osha.gov/dte/library/electrical/electrical_10.gif
We’ll talk moreaboutgroundinglater in the semesterwhen wecover faults
3
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
3BS
2 2 2, , ,3 1 1
( 3 )
3B LL B LN B LN
BB B B
V V VZ
S S S
Exactly the same impedance bases as with single phase!
4
Three Phase Per Unit, cont'd
4. Calculate the current base, IB
5. Convert actual values to per unit
3 1 13 1B B
, , ,
3I I
3 3 3B B B
B LL B LN B LN
S S S
V V V
Exactly the same current bases as with single phase!
5
Three Phase Per Unit Example
Solve for the current, load voltage and load power in the previous circuit, assuming a 3f power base of300 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 unitas before. Note thesystem is exactly thesame!
6
3f Per Unit Example, cont'd
L
2*
1.0 00.22 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.
LL L L
G
Ij
VS V I
ZS
Again, analysis is exactly the same!
7
3f Per Unit Example, cont'd
LActual
ActualL
ActualG
MiddleB
ActualMiddle
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 MVAI 125 (same cur0 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
8
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
9
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 NewBasepu actual pu
OriginalBase NewBasepu puOriginalBase NewBase
BaseBase
NewBaseOriginalBase NewBaseBasepu puOriginalBase
Base
Z Z Z
V VZ
SS
SZ
S
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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?
1000.0369 0.0683 p.u.
54eX
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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)?
2
1000.10 0.0286 p.u.
350
2300.0286 15.1
100
eX
12
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 core13
3f Transformer Interconnections
D-Y Y-D
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Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
1, ,An AB A
an ab a
V V Ia a
V V I a
15
Y-Y Connection: 3f Detailed Model
16
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
1 1, ,AB AB A
ab ab a
V I Ia
V I a I a
18
D-D Connection: 3f Detailed Model
To use the per phase equivalent we need to usethe delta-wye load transformation
19
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.
20
D-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
21
D-Y Connection V/I Relationships
, 3 30
3030Hence 3 and 3
For current we get
1
13 30 30
31
303
AB ABan ab an
an
AnABab an
ABa AB
ab
A AB AB A
a A
V Va V V V
V a
VVV V
a a
II a I
I a
I I I I
a I
22
D-Y Connection: Per Phase Model
Note: Connection introduces a 30 degree phase shift!
Common for transmission/distribution step-down sincethere is a neutral on the low voltage side.
Even if a = 1 there is a sqrt(3) step-up ratio
23
Y-D Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a phase shift of -30 degrees.
24
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
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
26
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 toprevent inadvertent "loop flow" and to prevent line overloads.
27Image Source: en.wikipedia.org/wiki/Quadrature_booster#/media/File:Qb-3ph.svg
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
28
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!
29
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
30
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
31
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
32
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
33