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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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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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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)