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2/28/2013
1
23: Economic Dispatch II Text: 11.7 – 11.12
ECEGR 451
Power Systems
1 Dr. Henry Louie Dr. Henry Louie 2
Topics
• Example
• Penalty Factor Calculation
• Example
Dr. Henry Louie 3
Example
• Consider the system without generator limits
• Find the optimal generation for each generator
and the power loss in the transmission line
1 G1
2 G2
2
L G2
D1
D2
IC 0.007P 4.1 $/MWh
IC 0.007P 4.1 $/MWh
P 0.001 P 50 MW
P 300 MW
P 50 MW
1 2
Dr. Henry Louie 4
Example
• First find
now find the penalty factor
L
G2
P
P
1
2
L
L
Dr. Henry Louie 5
Example
• First find
now find the penalty factor
• At the solution, L1xIC1 = L2xIC2 = l
LG2 G2
G2
P0.002 P 50 0.002P 0.1
P
1
2
G2
L 1
1L
1.1 0.002P
Dr. Henry Louie 6
Example
• We then have:
• According to the optimal dispatch rule:
G1
G2
G2
1 0.007P 4.1
10.007P 4.1
1.1 0.002P
l
l
1 G1
2 G2
1
2
G2
IC 0.007P 4.1 $/MWh
IC 0.007P 4.1 $/MWh
L 1
1L
1.1 0.002P
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Dr. Henry Louie 7
Example
• Start with a guess for l, and update if needed (depending on the corresponding generation values)
it is easier to rewrite the equations as:
G1
G2
4.1P
0.007
1.1 4.1P
0.007 0.002
l
l
l
G1
G1
G2
1 0.007P 4.1
10.007P 4.1
1.1 0.002P
l
l
Dr. Henry Louie 8
Example
• Start with a guess of l = 5.0
G1
G2
L
P
P
P
Dr. Henry Louie 9
Example
• Start with a guess of l = 5.0
• Power to the load is 128.6 + 82.4 – 1 = 210 MW (the load is 350 MW)
• We need more generation to meet the load, so increase l to 6.0
G1
G2
2
L
5.0 4.1P 128.6
0.007
1.1 5 4.1P 82.4
0.007 0.002 5
P 0.001 82.4 50 1.0
Example
• l = 6.0, what are the corresponding power and loss?
Dr. Henry Louie 10
G1
G2
L
P
P
P
Example
• l = 6.0, what are the corresponding power and loss?
• Power to the load is 271.4 + 131.6 – 6.7 = 396.3 MW (the load is 350 MW)
• Now there is too much generation; we need to decrease l
Dr. Henry Louie 11
G1
G2
2
L
6 4.1P 271.4
0.007
1.1 6 4.1P 131.6
0.007 0.002 6
P 0.001 131.6 50 6.7
Example
• After some more iterations, we find that l = 5.695 PG1 = 227.72 MW
PG2 = 117.65 MW
PL = 4.58 MW
• This is close enough to the exact answer
• Even though the generators have the same IC, due to losses, it is cheaper to have generator 1 supply more of the load
Dr. Henry Louie 12
1 2
300 MW 50 MW
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Dr. Henry Louie 13
Calculation of Penalty Factors
• What if we do not have a function relating losses to generation?
• The goal then is to find
• There is a problem, we haven’t found an expression for relating the losses in the system to the generator outputs
• Approach:
we know how to relate a change in angle to a change in real power injected
we know how to relate real power injected to losses
we can develop a relationship between angles and losses
L L
G2 Gm
P P, ,
P P
Dr. Henry Louie 14
Calculation of Penalty Factors
• Losses can be expressed as (with m gens, n buses)
• for each k = 2,…,n
q1 is constant and equal to zero (slack bus) note that we would expect the voltage magnitudes
to also change, but we are assuming that this change is small and hence the voltage magnitudes are assumed to be constant
we will come back to this expression later
m n n
L Gi Di ii 1 i 1 i 1
P P P P
L 1 m m 1 n
k k k k k
P P P P P
q q q q q
Assume: buses 1 to m are generators
Dr. Henry Louie 15
Calculation of Penalty Factors
• For any given q with elements (q2,…,qn) we explicitly find from the power flow equations
noting that
and for a fixed load
• We can find an alternate expression for recalling
• using the chain rule and the above relations
i
k
P
q
Gi i DiP P P
Gii
k k
PP
q q
L G L GL 2 m
k G2 k Gm k
P PP P P...
P Pq q q
P P
L
k
P
q
L L G L G2 GmP P P P ,...,P P
Dr. Henry Louie 16
Calculation of Penalty Factors
• This should be done for k = 2,…,n
• Which is subtracted from the expression on slide 14
to get, for each k:
L G L GL 2 m
k G2 k Gm k
P PP P P
P Pq q q
P P
1 2 L m L m 1 n
k k G2 k Gm k k
P P P P P P P1 1 0
P Pq q q q q
L 1 m m 1 n
k k k k k
P P P P P
q q q q q
Dr. Henry Louie 17
Calculation of Penalty Factors
• We can concatenate the equations for each k into a matrix
to get:
L
G2
2 m n 1
2 2 2 2
L
Gm
2 m n 1
n n n n
P1
PP P P P
... ...
P1
PP P P P
... ... 1
1
q q q q
q q q q
1 2 L m L m 1 n
k k G2 k Gm k k
P P P P P P P1 ... 1 ... 0
P Pq q q q q
Dr. Henry Louie 18
Calculation of Penalty Factors
L
G2
2 m n 1
2 2 2 2
L
Gm
2 m n 1
n n n n
P1
PP P P P
P1
PP P P P
1
1
q q q q
q q q q
11 i 1i 1 i 1i 1 i
i
PV V G sin B cosq q q q
q
transpose of
from the Jacobian in
the N-R method
q
P
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Dr. Henry Louie 19
Calculation of Penalty Factors
• We can then solve for the penalty factors:
L
G2
2 m n 1
2 2 2 2
L
Gm
2 m n 1
n n n n
P1
PP P P P
P1
PP P P P
1
1
q q q q
q q q q
Dr. Henry Louie 20
Outline for Optimal Dispatch
• Assume that none of the generators are at their limits
1. Start with a set of PDi, i = 2,…,n and an initial set of PGi, i = 2,…,m and use a power flow problem solver to find q1,…, qn (we can use the real power portion of a decoupled N-R, since we are assuming |Vi| for i = 1,…,n are known
2. We now find the power injected to the slack bus and then the generator power (P1 = PG1 – PD1)
Dr. Henry Louie 21
Outline for Optimal Dispatch
3. We now find for i = 2,…,n, and using we
calculate for i= 2,…,m. Taking the reciprocal
we can find the penalty factors, L2,…,Lm (we
already know L1 = 1)
4. We have all the PGi and penalty factors so we can compute the LixICi for each generator. If they are all equal, then we have the optimal dispatch. If not, we change PGi, i = 2,…,m
1
i
P
q
T
q
P
L
Gi
P1
P
Dr. Henry Louie 22
Example
• Consider the system:
• V1 = 1.0 at 0 deg.
• |V2| = 1.0
• |V3| = 1.0
• PD3 = 3.0 1 2 3
zL1 zL2
1 2 Gi
L1
L2
IC IC 4.1 0.7P
z 0.0099 j0.099
z 0.0198 j0.198
Note: Bus 3 voltage is known, but Q is not, so treat as PV bus.
Dr. Henry Louie 23
Example
• Consider the system:
• V1 = 1.0 at 0 deg.
• |V2| = 1.0
• |V3| = 1.0
• PD3 = 3.0 1 2 3
zL1 zL2
bus
j j
j j
j j j
1 10 0 1 10
0 0.5 5.0 0.5 5.0
1 10 0.5 5.0 1.5 15
Y
1 2 Gi
L1
L2
IC IC 4.1 0.7P
z 0.0099 j0.099
z 0.0198 j0.198
Note error in book
Dr. Henry Louie 24
Example
• The power injected into bus 1 is
• Find the equations for the power injected into bus 2 and bus 3
1 13 13P 1 cos 10sinq q q
2 23 23
3 31 31 32 32
P 0.5 0.5cos 5sin
P 1.5 cos 10sin 0.5cos 5sin
q q
q q q q
q
q
1 2 3 zL1 zL2
bus
j j
j j
j j j
1 10 0 1 10
0 0.5 5.0 0.5 5.0
1 10 0.5 5.0 1.5 15
Y
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Dr. Henry Louie 25
Example
• Losses are then:
3
L i 13 23i 1
P P 3 2cos cosq q
q
1 2 3 zL1 zL2
1 13 13
2 23 23
3 31 31 32 32
P 1 cos 10sin
P 0.5 0.5cos 5sin
P 1.5 cos 10sin 0.5cos 5sin
q q
q q
q q q q
q
q
q
bus
j j
j j
j j j
1 10 0 1 10
0 0.5 5.0 0.5 5.0
1 10 0.5 5.0 1.5 15
Y
recall that sin(x–y) = -sin(y-x)
cos(x-y) = cos(y-x)
Dr. Henry Louie 26
Example
• We also need to find
• Find the partial derivatives of
with respect to q2, q3
T
P
q
1 13 13
2 23 23
3 31 31 32 32
P 1 cos 10sin
P 0.5 0.5cos 5sin
P 1.5 cos 10sin 0.5cos 5sin
q q
q q
q q q q
q
q
q
Dr. Henry Louie 27
Example
• Resulting in:
223 23
2
223 23
3
332 32
2
331 31 32 32
3
P0.5sin 5cos
P0.5sin 5cos
P0.5sin 5cos
Psin 10cos 0.5sin 5cos
q qq
q qq
q qq
q q q qq
Example
• Now for the partial derivatives of:
• the system of equations is then:
Dr. Henry Louie 28
1
2
113 13
3
P0
Psin 10cos
q
q qq
32
L2 2
G2 1
323
3 3
PPP 0
1P P
PP1
q q
qq q
1 13 131 cos 10sinP q q q
Example
• Here we know:
• and we need to find the penalty factor
Dr. Henry Louie 29
32
L2 2
G2 1
323
3 3
PPP 0
1P P
PP1
q q
qq q
Dr. Henry Louie 30
Example
• Recall that
• It is simple to solve for
1 L2
G2
PL 1
P
L
23 23 32 32
G2
23 23 31 31 32 32 13 13
P10.5sin 5cos 0.5sin 5cos 0
P0.5sin 5cos sin 10cos 0.5sin 5cos sin 10cos
1
q q q q
q q q q q q q q
1 L2
G2
PL 1
P
L23 23 32 32
G2
32 32L
G2 23 23
23 232
32 32
P(0.5sin 5cos )(1 ) ( 0.5sin 5cos ) 0
P
0.5sin 5cosP1
P 0.5sin 5cos
0.5sin 5cosL
0.5sin 5cos
q q q q
q q
q q
q q
q q
Eqn 11.85 in the text appears to be incorrect
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Dr. Henry Louie 31
Example
• We are now ready to begin the solution steps previously outlined
Dr. Henry Louie 32
Example
• Try PG2 = 1.5 as an initial guess
• Step 1: solve the power flow problem:
q
q
q
23
13
31
17.19
8.96
8.96
2
3
8.23
8.96
q
q
We will need these values to compute the penalty factor
and compute
Dr. Henry Louie 33
Example
• Step 2: find the power from the slack bus generator
1 13 13P 1 cos 10sin 1.57q q
Dr. Henry Louie 34
Example
• Step 3: find the penalty factors
we know the penalty factor at the slack bus is 1
to find L2, use (for this example):
• Step 4: compute the LixICi for each generator
1
2
1 1
2 2
IC 5.198
IC 5.150
L xIC 5.198
L xIC 5.479
23 232
32 32
0.5sin 5cosL 1.06
0.5sin 5cos
q q
q q
Dr. Henry Louie 35
Example
• Should we stop here?
• No, the incremental costs are not equal
• We should decrease PG2 to drive down L2xIC2
• Now try PG2 = 1.3 and repeat steps 1 – 4.
1 1
2 2
L xIC 5.198
L xIC 5.479
Dr. Henry Louie 36
Example
• This yields
• Should we stop here?
• no, the incremental costs are not equal
• We should increase PG2 to increase L2xIC2
• The iterations continue until we reach PG1 = 1.734 and PG2 = 1.331, corresponding to
1 1
2 2
L xIC 5.335
L xIC 5.283
1 1
2 2
L xIC 5.313
L xIC 5.313
Note: L2 is updated after each iteration
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Dr. Henry Louie 37
Economics Summary
• The economic dispatch (ED) problem comes in several flavors depending on
loss considerations and generator limits
• The goal is to minimize the fuel costs needed to serve the load
• We use a solution approach based on the incremental costs of generators
• The solution approaches presented have used approximate system models (voltage, reactive power, line constraints, etc have been ignored)
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