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Transcript
ECE 476
POWER SYSTEM ANALYSIS
Lecture 3
Three Phase, Power System Operation
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Reading and Homework
•
•
For lecture 3 please be reading Chapters 1 and 2
For lectures 4 through 6 please be reading Chapter 4
–
•
we will not be covering sections 4.7, 4.11, and 4.12 in
detail
HW 1 is 2.7, 12, 21, 26; due Thursday 9/4
1
Balanced 3 Phase () Systems

A balanced 3 phase () system has
three voltage sources with equal magnitude, but with an
angle shift of 120
equal loads on each phase
equal impedance on the lines connecting the generators to
the loads


Bulk power systems are almost exclusively 3
Single phase is used primarily only in low voltage,
low power settings, such as residential and some
commercial
2
Balanced 3 -- No Neutral Current
I n  I a  Ib  I c
V
In 
(10  1   1  
Z
*
*
*
*
S  Van I an
 Vbn I bn
 Vcn I cn
 3 Van I an
3
Advantages of 3 Power




Can transmit more power for same amount of wire
(twice as much as single phase)
Torque produced by 3 machines is constrant
Three phase machines use less material for same
power rating
Three phase machines start more easily than single
phase machines
4
Three Phase - Wye Connection

There are two ways to connect 3 systems
Wye (Y)
Delta ()
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
5
Wye Connection Line Voltages
Vca
Vcn
Vab
-Vbn
Van
Vbn
Vbc
Vab
(α = 0 in this case)
 Van  Vbn  V (1  1  120

3 V   30
Vbc

3 V   90
Vca

3 V   150
Line to line
voltages are
also balanced
6
Wye Connection, cont’d


Define voltage/current across/through device to be
phase voltage/current
Define voltage/current across/through lines to be
line voltage/current
VLine  3 VPhase 130  3 VPhase e
j
6
I Line  I Phase
S3
*
 3 VPhase I Phase
7
Delta Connection
For the Delta
phase voltages equal
line voltages
Ica
For currents
Ia  I ab  I ca
Ic

Ib
Ibc
Iab
Ia
3 I ab   
I b  I bc  I ab
Ic  I ca  I bc
S3 
*
3 VPhase I Phase
8
Three Phase Example
Assume a -connected load is supplied from a 3
13.8 kV (L-L) source with Z = 10020W
Vab  13.80 kV
Vbc  13.8 0 kV
Vca  13.80 kV
13.80 kV
I ab 
 138  20 amps
 W
I bc  138  140 amps
I ca  1380 amps
9
Three Phase Example, cont’d
I a  I ab  I ca  138  20  1380
 239  50 amps
I b  239  170 amps I c  2390 amps
*
S  3  Vab I ab
 3  13.80kV  138 amps
 5.7 MVA
 5.37  j1.95 MVA
pf  cos 20   lagging
10
Delta-Wye Transformation
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with ZY  Z 
3
2) Δ-connected sources can be replaced by
VLine
Y-connected sources with Vphase 
330
11
Delta-Wye Transformation Proof
From the  side we get
Vab Vca
Vab  Vca
Ia 


Z Z
Z
Hence
Vab  Vca
Z 
Ia
12
Delta-Wye Transformation, cont’d
From the Y side we get
Vab  ZY ( I a  I b )
Vca  ZY ( I c  I a )
Vab  Vca  ZY (2 I a  I b  I c )
Since
Ia  I b  I c  0  I a   I b  I c
Hence
Vab  Vca  3 ZY I a
3 ZY
Vab  Vca

 Z
Ia
Therefore
ZY
1
 Z
3
13
Three Phase Transmission Line
14
Per Phase Analysis


Per phase analysis allows analysis of balanced 3
systems with the same effort as for a single phase
system
Balanced 3 Theorem: For a balanced 3 system
with
–
–
All loads and sources Y connected
No mutual Inductance between phases
15
Per Phase Analysis, cont’d

Then
–
–
–
All neutrals are at the same potential
All phases are COMPLETELY decoupled
All system values are the same sequence as sources. The
sequence order we’ve been using (phase b lags phase a
and phase c lags phase a) is known as “positive”
sequence; later in the course we’ll discuss negative and
zero sequence systems.
16
Per Phase Analysis Procedure
To do per phase analysis
1. Convert all  load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)
5. If necessary, go back to original circuit to determine
line-line values or internal  values.
17
Per Phase Example
Assume a 3, Y-connected generator with Van = 10
volts supplies a -connected load with Z = -jW
through a transmission line with impedance of j0.1W
per phase. The load is also connected to a
-connected generator with Va”b” = 10 through a
second transmission line which also has an impedance
of j0.1W per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each generator, SY and
S
18
Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
19
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
1
'
'
'
(Va  10)(10 j )  Va (3 j )  (Va 
   j  
3
20
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
1
'
'
'
(Va  10)(10 j )  Va (3 j )  (Va 
   j  
3
10
(10 j 
60)  Va' (10 j  3 j  10 j )
3
Va'  0.9  volts
Vb'  0.9  volts
Vc'  0.9 volts
'
Vab
 1.56 volts
21
Per Phase Example, cont’d
*
'
 Va  Va 
Sygen  3Va I a*  Va 
  5.1  j 3.5 VA
 j 0.1 
"

" Va
Sgen  3Va 
' *
 Va
  5.1  j 4.7 VA
 j 0.1 
22
Power System Operations Overview



Goal is to provide an intuitive feel for power system
operation
Emphasis will be on the impact of the transmission
system
Introduce basic power flow concepts through small
system examples
23
Power System Basics




All power systems have three major components:
Generation, Load and Transmission/Distribution.
Generation: Creates electric power.
Load: Consumes electric power.
Transmission/Distribution: Transmits electric power
from generation to load.
–
–
Lines/transformers operating at voltages above 100 kV are
usually called the transmission system. The transmission
system is usually networked.
Lines/transformers operating at voltages below 100 kV are
usually called the distribution system (radial).
24
Small PowerWorld Simulator Case
Load with
green
arrows
indicating
amount
of MW
flow
Bus 2
20 MW
-4 MVR
Bus 1
1.00 PU
204 MW
102 MVR
1.00 PU
106 MW
0 MVR
150 MW AGC ON
116 MVR AVR ON
-14 MW
-34 MW
10 MVR
4 MVR
34 MW
-10 MVR
Home Area
Used
to control
output of
generator
-20 MW
4 MVR
100 MW
Note the
power
balance at
each bus
14 MW
-4 MVR
1.00 PU
Bus 3
102 MW
51 MVR
150 MW AGC ON
37 MVR AVR ON
Direction of arrow is used to indicate
direction of real power (MW) flow
25
Power Balance Constraints




Power flow refers to how the power is moving
through the system.
At all times in the simulation the total power
flowing into any bus MUST be zero!
This is know as Kirchhoff’s law. And it can not be
repealed or modified.
Power is lost in the transmission system.
26
Basic Power Control



Opening a circuit breaker causes the power flow to
instantaneously(nearly) change.
No other way to directly control power flow in a
transmission line.
By changing generation we can indirectly change
this flow.
27
Transmission Line Limits



Power flow in transmission line is limited by
heating considerations.
Losses (I2 R) can heat up the line, causing it to sag.
Each line has a limit; Simulator does not allow you
to continually exceed this limit. Many utilities use
winter/summer limits.
28
Overloaded Transmission Line
29
Interconnected Operation


Power systems are interconnected across large
distances. For example most of North America east
of the Rockies is one system, with most of Texas
and Quebec being major exceptions
Individual utilities only own and operate a small
portion of the system, which is referred to an
operating area (or an area).
30
Operating Areas



Transmission lines that join two areas are known as
tie-lines.
The net power out of an area is the sum of the flow
on its tie-lines.
The flow out of an area is equal to
total gen - total load - total losses = tie-flow
31
Area Control Error (ACE)



The area control error is the difference between the
actual flow out of an area, and the scheduled flow.
Ideally the ACE should always be zero.
Because the load is constantly changing, each utility
must constantly change its generation to “chase” the
ACE.
32
Automatic Generation Control


Most utilities use automatic generation control
(AGC) to automatically change their generation to
keep their ACE close to zero.
Usually the utility control center calculates ACE
based upon tie-line flows; then the AGC module
sends control signals out to the generators every
couple seconds.
33
Three Bus Case on AGC
Bus 2
-40 MW
8 MVR
40 MW
-8 MVR
Bus 1
1.00 PU
266 MW
133 MVR
1.00 PU
101 MW
5 MVR
150 MW AGC ON
166 MVR AVR ON
-39 MW
-77 MW
25 MVR
12 MVR
78 MW
-21 MVR
Home Area
Generation
is automatically
changed to match
change in load
100 MW
39 MW
-11 MVR
Bus 3
1.00 PU
133 MW
67 MVR
250 MW AGC ON
34 MVR AVR ON
Net tie flow is
close to zero
34
Generator Costs





There are many fixed and variable costs associated
with power system operation.
The major variable cost is associated with
generation.
Cost to generate a MWh can vary widely.
For some types of units (such as hydro and nuclear)
it is difficult to quantify.
For thermal units it is much easier. These costs will
be discussed later in the course.
35
Economic Dispatch


Economic dispatch (ED) determines the least cost
dispatch of generation for an area.
For a lossless system, the ED occurs when all the
generators have equal marginal costs.
IC1(PG,1) = IC2(PG,2) = … = ICm(PG,m)
36
Power Transactions



Power transactions are contracts between areas to
do power transactions.
Contracts can be for any amount of time at any
price for any amount of power.
Scheduled power transactions are implemented by
modifying the area ACE:
ACE = Pactual,tie-flow - Psched
37
100 MW Transaction
Bus 2
8 MW
-2 MVR
-8 MW
2 MVR
Bus 1
1.00 PU
225 MW
113 MVR
1.00 PU
0 MW
32 MVR
150 MW AGC ON
138 MVR AVR ON
-92 MW
-84 MW
27 MVR
30 MVR
85 MW
-23 MVR
Home Area
93 MW
-25 MVR
Bus 3
Scheduled Transactions
100.0 MW
Scheduled
100 MW
Transaction from Left to Right
100 MW
1.00 PU
113 MW
56 MVR
291 MW AGC ON
8 MVR AVR ON
Net tie-line
flow is now
100 MW
38
Security Constrained ED
Transmission constraints often limit system
economics.
 Such limits required a constrained dispatch in
order to maintain system security.
 In three bus case the generation at bus 3 must
be constrained to avoid overloading the line
from bus 2 to bus 3.

39
Security Constrained Dispatch
Bus 2
-22 MW
4 MVR
22 MW
-4 MVR
Bus 1
1.00 PU
357 MW
179 MVR
1.00 PU
0 MW
37 MVR
100%
194 MW OFF AGC -142 MW
49 MVR
232 MVR AVR ON
145 MW 100%
-37 MVR
Home Area
Bus 3
Scheduled Transactions
100.0 MW
-122 MW
41 MVR
100 MW
124 MW
-33 MVR
1.00 PU
179 MW
89 MVR
448 MW AGC ON
19 MVR AVR ON
Dispatch is no longer optimal due to need to keep line
from bus 2 to bus 3 from overloading
40
Multi-Area Operation



If Areas have direct interconnections, then they may
directly transact up to the capacity of their tie-lines.
Actual power flows through the entire network
according to the impedance of the transmission
lines.
Flow through other areas is known as “parallel
path” or “loop flows.”
41
Seven Bus Case: One-line
System has
three areas
44 MW
-42 MW
-31 MW
0.99 PU
3
1.05 PU
1
106 MW -37 MW
AGC ON
62 MW
79 MW
2
40 MW
20 MVR
Top Area Cost
8029 $/MWH
1.00 PU
-32 MW
Case Hourly Cost
16933 $/MWH
32 MW
80 MW
30 MVR
4
110 MW
40 MVR
38 MW
-61 MW
1.04 PU
31 MW
-77 MW
5
-39 MW
40 MW
94 MW
AGC ON
Area top
has five
buses
-14 MW
1.01 PU
130 MW
40 MVR
168 MW AGC ON
-40 MW
1.04 PU
6
Area left
has one
bus
20 MW
-20 MW
40 MW
1.04 PU
20 MW
200 MW
0 MVR Left Area Cost
4189 $/MWH
200 MW AGC ON
-20 MW
7
200 MW
Right Area Cost
0 MVR
4715 $/MWH
201 MW AGC ON
Area right has one
bus
42
Seven Bus Case: Area View
Top
40.1 MW
0.0 MW
Area Losses
7.09 MW
-40.1 MW
0.0 MW
System has
40 MW of
“Loop Flow”
Left
Area Losses
0.33 MW
Right
40.1 MW
0.0 MW
Actual
flow
between
areas
Scheduled
flow
Area Losses
0.65 MW
Loop flow can result in higher losses
43
Seven Bus - Loop Flow?
Top
4.8 MW
0.0 MW
-4.8 MW
0.0 MW
Left
Area Losses
-0.00 MW
100 MW Transaction
between Left and Right
Area Losses
9.44 MW
Right
104.8 MW
100.0 MW
Note that
Top’s
Losses have
increased
from
7.09MW to
9.44 MW
Area Losses
4.34 MW
Transaction has
actually decreased
the loop flow
44
Pricing Electricity





Cost to supply electricity to bus is called the
locational marginal price (LMP)
Presently some electric makets post LMPs on the
web
In an ideal electricity market with no transmission
limitations the LMPs are equal
Transmission constraints can segment a market,
resulting in differing LMP
Determination of LMPs requires the solution on an
Optimal Power Flow (OPF)
45
3 BUS LMPS - OVERLOAD IGNORED
Bus 2
Gen 2’s
cost
is $12
per
MWh
60 MW
60 MW
Bus 1
10.00 $/MWh
0 MW
10.00 $/MWh
120 MW
120%
180 MW
0 MW
Gen 1’s
cost
is $10
per
MWh
60 MW
Total Cost
1800 $/hr
120%
120 MW
60 MW
10.00 $/MWh
Bus 3
180 MW
0 MW
Line from Bus 1 to Bus 3 is over-loaded; all
buses have same marginal cost
46
LINE OVERLOAD ENFORCED
Bus 2
20 MW
20 MW
Bus 1
10.00 $/MWh
60 MW
12.00 $/MWh
100 MW
80%
100%
120 MW
0 MW
80 MW
Total Cost
1921 $/hr
80%
100%
100 MW
80 MW
14.01 $/MWh
Bus 3
180 MW
0 MW
Line from 1 to 3 is no longer overloaded, but now
the marginal cost of electricity at 3 is $14 / MWh
47