ECE 476
Power System Analysis
Lecture 18: LMP Markets,
Symmetrical Faults and Components
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Read Chapter 8
• HW 8 is due now
• HW 9 is 7.6, 7.13, 7.19, 7.28, 8.4; will be covered by
an in-class quiz on Nov 5
• Second exam is Thursday Nov. 12 during class.
Closed book, closed notes, one note sheet and
calculators allowed
• Abbott Power Plant tour is Tuesday Nov. 10 during
class (we’ll be meeting at Abbott though)
1
Generator Costs
• As discussed in the economic dispatch section,
generators have incremental costs that vary according
to their output
• If an OPF is cost-based, then the costs curves can be
expressed either as continuous functions or in
piecewise linear blocks
• With a piecewise linear approach there can be a no
load cost plus an incremental energy cost for blocks
(first 50 MW at $20/MWh, second 50 MWs at
$25/MWh)
2
Generator Offers
• Traditionally utility generator costs were based on the
actual cost of operating the generator
• As power industry has restructured and become more
competitive, there has been a move to price-based
curves
–
•
Competitive business do not need to sell their products at
their actual production cost (if they did they might not stay
in business long)
In many power markets the generators submit
“offers” to sell power
–
Offers can either be cost-based or price-based
3
Electricity Markets
• Worldwide there are a large number of electricity
markets, each with
their own rules
• Electricity markets
provide for competition
–
Market might have
many competitive companies owning a portfolio of
individual generators
• A third party (e.g., an independent system operator
[ISO]) runs the market to get enough electricity to
keep the lights on
Image: www.ferc.gov/market-oversight/mkt-electric/overview.asp
4
Electricity Markets, cont.
• Electricity markets trade a number of different
commodities, with MWh being the most important
• Commonly a market has two settlement periods:
day ahead and real-time
–
–
Day Ahead: Generators (and possibly loads) submit
offers for the next day; OPF is used to determine who
gets dispatched based upon forecasted conditions.
Results are financially binding
Real-time: Modifies the day ahead market based upon
real-time conditions.
5
Payment
• Generators are only paid if they are selected to run,
otherwise they stay off-line
• In most markets generators are not paid their offer,
rather they are paid the LMP at their bus, the loads pay
the LMP
• At the residential/commercial level the LMP costs are
usually not passed on directly to the end consumer.
Rather, they these consumers typically pay a fixed rate.
• LMPs may differ across a system due to transmission
system “congestion.”
6
MISO LMP Contour: Oct 23, 2015,
8:05 am CDT
Image: https://www.misoenergy.org/LMPContourMap/MISO_All.html
7
Some Electricity Market Challenges
• Running a real-time electricity market can be quite
challenging
–
–
the total generation must always match the total load plus
losses; the load is continuously varying; generators can fail
the transmission limited capacity; transmission line and
transformer flows must be within their limits; bus voltages
must be within their limits
• Goal of the market is to allow for competition, so the
generator owners are seeking to maximize their
profits
• There is a need to prevent generators from exercising
market power
8
Profit Maximization: 30 Bus Example
• For the below 30 bus case assume the generators at
buses 13 and 23 have true costs of $30/MWh
• If they each offer at $30/MWh LMPs are shown; their
net profit is 35*0.38 = $13.3/hr and 50*0.38 = $19/hr
43.56 MW
58.22 MW
slack
A
1
96%
2
MV A
1.000
18
15
A
60%
MVA
A
87%
A
A
A
A
A
MVA
MVA
MVA
MVA
14
A
A
A
4
A
4 MW
MVA
MVA
MVA
A
57%
A
MVA
12
MVA
7
8
19
30.38 $/MWh
MVA
MVA
3
A
A
MV A
35.00 MW
MVA
28
MVA
6
A
13
5
75%
MVA
A
A
A
59%
MVA
A
MVA
MVA
MVA
9
11
A
16
8 MW
This is a
rather low
profit!
17
13 MW
MVA
A
9 MW
A
MVA
MVA
9 MW
A
20
56%
26
A
17 MW
10
18.00 MW
MVA
A
MVA
MVA
23
A
77%
50.00 MW
MV A
25
22
A
21
24
A
90%
MVA
A
32.00 MW
30.38 $/MWh
14 MW
MV A
94%
MV A
A
MVA
29
27
30
A
A
MVA
A
MVA
MVA
9
Profit Maximization: 30 Bus Example
• Now assume the generator at 13 offers $50/MWh
• The result is shown, which is good for the
generator at 23 but bad for the one at 13
The bus 13
gen is off,
1.000
but the
one at
23 has a
profit of
42.47*20
=$849.94/hr
46.47 MW
62.54 MW
slack
A
1
72%
2
MVA
18
15
A
61%
MVA
A
A
59%
A
A
A
A
MVA
MVA
MVA
MVA
MVA
28
14
A
A
48.40 $/MWh
MVA
MVA
3
A
A
A
4
A
MVA
A
MVA
20 MW
MVA
A
100%
MVA
7
8
19
0.00 MW
MVA
12
A
MVA
6
95%
13
MVA
5
MV A
A
A
A
MVA
A
MVA
MVA
MVA
9
11
A
16
7 MW
17
12 MW
MVA
A
18 MW
A
MVA
58%
MVA
16 MW
A
20
68%
26
A
21 MW
10
35.08 MW
MVA
A
MVA
MVA
23
A
91%
50.00 MW
MV A
25
22
A
21
24
42.47 $/MWh
7 MW
A
MVA
MVA
A
43.00 MW
78%
A
66%
MV A
MVA
29
27
30
A
A
MVA
A
MVA
MVA
10
Profit Maximization: 30 Bus Example
• If they both offer $200/MWh they end up making a
lot of profit!
The bus 13
gen makes
1.000
$4472/hr
and the one
at 23
$687/hr
They have
market
power!
46.87 MW
67.00 MW
slack
A
1
59%
2
MVA
18
15
A
59%
MVA
A
A
A
A
A
A
MVA
MVA
MVA
MVA
MVA
28
14
A
66%
A
200.00 $/MWh
MVA
MVA
3
A
A
MVA
A
4
A
20 MW
A
MVA
MVA
A
100%
MVA
7
8
19
26.31 MW
MVA
12
MV A
5
6
MVA
A
13
MVA
A
A
A
MVA
A
MVA
MVA
MVA
9
8 MW
11
A
16
17
15 MW
MVA
A
16 MW
A
MVA
52%
MVA
15 MW
A
75%
26
A
24 MW
10
45.00 MW
20
MV A
A
MVA
MVA
23
A
86%
5.00 MW
MV A
25
22
A
21
24
59%
167.40 $/MWh
10 MW
A
89%
MVA
47.00 MW
A
100%
MV A
A
MVA
MVA
29
27
30
A
A
MVA
A
MVA
MVA
Market rules are set to minimize such situations
11
Generator Short Circuit Example
• Returning to balanced three-phase faults
• 500 MVA, 20 kV, 3 is operated with an internal
voltage of 1.05 pu. Assume a solid 3 fault occurs
on the generator's terminal and that the circuit
breaker operates after three cycles. Determine the
fault current. Assume
X d"  0.15,
X d'  0.24,
Td"  0.035 seconds, Td'
X d  1.1 (all per unit)
 2.0 seconds
TA  0.2 seconds
12
Generator S.C. Example, cont'd
Substituting in the values
1   t 2.0 
1  1

1.1   0.24  1.1  e
I ac (t )  1.05 

t
 1  1  e  0.035

 0.15 0.24 

I ac (0)  1.05
 7 p.u.
0.15
I base

500  106
 14,433 A I ac (0)  101,000 A
3
3 20  10
I DC (0)  101 kA  2 e
t
0.2
 143 k A I RMS (0)  175 kA
13
Generator S.C. Example, cont'd
Evaluating at t = 0.05 seconds for breaker opening
1   0.05 2.0 
1  1

1.1   0.24  1.1  e
I ac (0.05)  1.05 


0.05
 1  1  e

0.035
 0.15 0.24 

I ac (0.05)  70.8 kA
I DC (0.05)  143  e
0.05
0.2
kA  111 k A
I RMS (0.05)  70.82  1112  132 kA
14
Network Fault Analysis Simplifications
• To simplify analysis of fault currents in networks
we'll make several simplifications:
–
–
–
–
–
Transmission lines are represented by their series
reactance
Transformers are represented by their leakage reactances
Synchronous machines are modeled as a constant voltage
behind direct-axis subtransient reactance
Induction motors are ignored or treated as synchronous
machines
Other (nonspinning) loads are ignored
15
Network Fault Example
For the following network assume a fault on the
terminal of the generator; all data is per unit
except for the transmission line reactance
generator has 1.05
terminal voltage &
supplies 100 MVA
with 0.95 lag pf
Convert to per unit: X line
19.5

 0.1 per unit
2
138
100
16
Network Fault Example, cont'd
Faulted network per unit diagram
To determine the fault current we need to first estimate
the internal voltages for the generator and motor
For the generator VT  1.05, SG  1.018.2
*
I Gen
1.018.2 


  0.952  18.2
 1.05 
E 'a  1.1037.1
17
Network Fault Example, cont'd
The motor's terminal voltage is then
1.050 - (0.9044 - j 0.2973)  j 0.3  1.00  15.8
The motor's internal voltage is
1.00  15.8  (0.9044 - j 0.2973)  j 0.2
 1.008  26.6
We can then solve as a linear circuit:
1.1037.1 1.008  26.6
If 

j 0.15
j 0.5
 7.353  82.9  2.016  116.6  j9.09
18
Fault Analysis Solution Techniques
• Circuit models used during the fault allow the
network to be represented as a linear circuit
• There are two main methods for solving for fault
currents:
–
–
Direct method: Use prefault conditions to solve for the
internal machine voltages; then apply fault and solve
directly
Superposition: Fault is represented by two opposing
voltage sources; solve system by superposition
• first voltage just represents the prefault operating point
• second system only has a single voltage source
19
Superposition Approach
Faulted Condition
Fault is represented
by two equal and
Exact Equivalent to Faulted Condition opposite voltage
sources, each with
a magnitude equal
to the pre-fault voltage
20
Superposition Approach, cont’d
Since this is now a linear network, the faulted voltages
and currents are just the sum of the pre-fault conditions
[the (1) component] and the conditions with just a single
voltage source at the fault location [the (2) component]
Pre-fault (1) component equal to the pre-fault
power flow solution
Obviously the
pre-fault
“fault current”
is zero!
21
Superposition Approach, cont’d
Fault (1) component due to a single voltage source
at the fault location, with a magnitude equal to the
negative of the pre-fault voltage at the fault location.
I g  I (1)  I g(2)
g
Im  I m(1)  I m(2)
(2)
(2)
I f  I (1)

I

0

I
f
f
f
22
Two Bus Superposition Solution
Before the fault we had E f  1.050,
I (1)  0.952  18.2 and I m(1)  0.952  18.2
g
Solving for the (2) network we get
Ef
1.050
(2)
Ig


  j7
j0.15
j0.15
E f 1.050
(2)
Im


  j 2.1
j0.5
j0.5
(2)
If
  j 7  j 2.1   j 9.1
This matches
what we
calculated
earlier
I g  0.952  18.2  j 7  7.35  82.9
23
Extension to Larger Systems
The superposition approach can be easily extended
to larger systems. Using the Ybus we have
Ybus V  I
For the second (2) system there is only one voltage
source so I is all zeros except at the fault location


 0 


I   I f 


 0 


However to use this
approach we need to
first determine If
24
Determination of Fault Current
Define the bus impedance matrix Z bus as
Z bus
 Z11
Then 

 Z n1
1
Ybus
V  Z busI
(2) 

V


1
 (2) 


Z1n  0
V

  2 
  I   

 f  

Z nn   0  V (2) 
n 1

  (2) 
Vn 
For a fault a bus i we get -If Zii  V f  Vi(1)
25
Determination of Fault Current
Hence
Vi(1)
If 
Zii
Where
Zii
driving point impedance
Zij (i  j )
transfer point imepdance
Voltages during the fault are also found by superposition
Vi  Vi(1)  Vi(2)
Vi(1) are prefault values
26