Power Control ED

8/3/2019 Power Control ED

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ELEN 460:Power System Control and Operation:Economic Dispatch (ED) 1

ELEN 460:Power System Control and Operation:

Economic Dispatch (ED)

Chanan Singh

Panida Jirutitijaroen




Outline

• Read chapter 11: 11.7-11.11.

• Homework: 11.8, 11.10, 11.10-11

• Overview of Economic Dispatch (ED)

• Economic Dispatch formulation• Case 1: No line losses, no generator limit.

– Example 11.8

• Case 2: No line losses, with generator limit.

• Case 3: With line losses, no generator limit. – Penalty factor

• Case 4: With line losses, with generator limit.




Overview of ED

• Consider cost of operation when selectinggenerating units to pick up the load.

• ED tries to minimize this cost.

• Assume that fuel-cost curve of each generatingunit is known and expressed in terms of theoutput power,

Where = Cost of generation for unit i

= Output power of unit i

dollars/hr2

GiiGiiiGiiPPPC

GiiPC

GiP




ED Formulation

• To minimize cost of operation,

• Subject to:

– Power flow equation

– Generation limit

– Voltage limit

– Line flow limit

m

i

GiiT PC C 1

0,,,32

11

GmGGloss

n

i

Di

m

i

GiGi PPPPPPP f

maxmin

GiGiGi PPP

max

ijij PP

maxmin

iii V V V

Assume bus 1 is slack bus, only m-1

independent power variable

Neglect line flow limitin this analysis

Neglect voltage limit dueto weak coupling effect




ED Simplified Formulation

• To minimize cost of operation,

• Subject to:

– Power flow equation

– Generation limit

m

i

GiiT PC C 1

0,,,32

11

GmGGloss

n

i

Di

m

i

GiGi PPPPPPP f

maxmin

GiGiGi PPP




Solution to ED

• The problem can categorized into nonlinearoptimization problem.

• Lagrangian of the problem is,

• is called Lagrange multiplier.

• Optimality condition:

0..

min

xgt s

x f

xg x f x L ,min

0

,

0,

x L

x

x L

**, xOptimalSolution




Next,

• We will consider 4 cases.

– Case 1: No line losses, no generator limit.

– Case 2: No line losses, with generator limit.

– Case 3: With line losses, no generator limit. – Case 4: With line losses, with generator limit.




Case 1: No Line Losses, No Generator Limit

• minimize

• Subject to:


• Optimal condition,

• Solution,

D

m

i

GiT T PPC C 1

~

00~

,,2,1,00

~

1

D

m

i

GiT

Gi

Gii

Gi

T

PPC

miP

PC

P

C

mi

P

PC IC

Gi

Giii ,,2,1

m

i

GiiT PC C 1

01

D

m

i

Gi PP




Case 1: Alternative Solution Procedure

• minimize

• Subject to:

• From this constraint, we can reduce number ofvariable by substitution to the objective function,

• The problem becomes unconstrained problem,to minimize

m

i

GiiT PC C

1

01

D

m

i

Gi PP

121 GmGG DGm PPPPP

121112211

GmGG DmGmmGGT PPPPC PC PC PC C




• Necessary condition:

• Let incremental cost,

• The solution is to have all ICs the same for allunits.

0

Gi

T

P

C

0

Gi

Gm

Gm

m

Gi

i

P

P

P

C

P

C

1,,2,1

mi

PC

PC

Gm

m

Gi

i

1,,2,1

miP

C

IC Gi

i

i

-1




Case 1: Solution Interpretation

• Incremental Cost is the slope of fuel-cost curve.

• Unit of IC is dollar per MWh.

• IC tells how much it costs to operate this unit for the next

MW power.• Example: two units with IC1 > IC2.

• This means that, for additional 1 MW power,

– Operating unit 1 cost more than operating unit 2.

– To minimize operation cost, it is reasonable to reduce poweroutput of unit 1 and increase that of unit 2.

– At optimal, additional cost from unit 1 and 2 should be the same.




Example 11.8

• Two generators with the following cost curve.

2

111101.045900 GGG PPPC 2

2222003.0432500 GGG PPPC

1

1

11

1

02.045G

G

G PP

PC IC

2

2

22

2

006.043G

G

G PP

PC IC

MW70021

DGG PPP

1121

700006.04302.045 GG PP IC IC

MW6.841 GP MW4.6152 GP

/MWh69.4621 IC IC




0 100 200 300 400 500 600 70042

44

46

48

50

52

54

56

58

60

Example 11.8

0 100 200 300 400 500 600 70042

44

46

48

50

52

54

56

58

60

P1 = 84.6 MW P2 = 615.4 MW

IC1 = IC2 = 46.69

1

2

1

2

3

++

= Load




Case 1: Solution Outline

• In general case with more than 2 generatingunits, we know only

– Total load

– IC curves of each unit• Iterative procedure:

Step 1: Pick initial

Step 2: Find corresponding power output of each unit

Step 3: If total power < load, increase and go backto step 2. Else, stop.




Case 2: No Line Losses, With Generator Limit

• Consider

• Assume that ALL UNITS are in operation.

• When some generating units hit their limit, wecan no longer operate all units at the same ICs.

maxmin

GiGiGi PPP

IC

Output Power

1

2

3

1,2,3 operate at lower limit

1 operate at lower limitIC2 = IC3

IC1= IC2 = IC3

3 operate at upper limitIC1 = IC2

2,3 operate atupper limit




Case 3: With Line Losses, No Generator Limit

• Minimize,

• Subject to power flow equation


m

i

GiiT PC C 1

0,,, 3211

GmGGloss

n

i Di

m

iGiGi PPPPPPP f

DGmGGloss

m

iGiT T PPPPPPC C ,,,

~

321




Case 3: Optimality Condition

• Lagrangian of the problem,

• Optimality condition,

00

~

,,3,2,010

~

00

~

1

1

11

1

Dloss

m

i

Gi

T

Gi

loss

Gi

Gii

Gi

T

G

G

G

T

PPPC

miP

P

P

PC

P

C

P

PC

P

C

DGmGGloss

m

i

GiT T PPPPPPC C ,,,~

32

1




Penalty Factor

• Rewrite optimality condition 2,

• Define “penalty factor”,

• From optimality condition 1,

we have

mi

P

PC

P

P Gi

Gii

Gi

loss

,,2,1

1

1

Gi

loss

i

P

P L

1

1

11 L

i L

1

11

G

G

P

PC




Case 3: Optimality Condition Revisit

• Optimality condition,

• Equivalently,

and

00

~

,,3,2,0

~

0

~

1

1

11

1

1

Dloss

m

i

GiT

Gi

Giii

Gi

T

G

G

G

T

PPPC

miP

PC LP

C

P

PC L

P

C

mi IC L ii ,,2,1

IC of unit i

01

Dloss

m

i

GiPPP




Case 3: Solution Outline,Penalty Factor Known

• Assume that the following is known. – Total load

– IC curves of each unit

– Penalty factor of each unit (from explicit loss

expression)

• Iterative procedure:Step 1: Pick initial

Step 2: Find corresponding power output of each unit

fromStep 3: If total power < load, increase and go back

to step 2. Else, stop.

mi IC L ii ,,2,1




Penalty Factor Calculation

• Consider power flow equation with loss,

• We can rewrite equation as follow:

• Differentiate with respect to , we have

• From,

0,,,32

11

GmGGloss

n

i

Di

m

i

GiGi PPPPPPP f

m

i

GiGmGGloss DG PPPPPPP2

321,,,

mi

P

P

P

P

Gi

loss

Gi

G ,,3,211

GiP

Gi

loss

i

P

P L

1

1mi

LP

P

iGi

G ,,3,21

1




Use of Jacobian Matrix to Find Penalty Factor

• From,

• Use as the intermediaries together with chainrule,

niP

P

PP

P

PP

P

P

P

P

PP

P

PP

P

PP

i

m

m

G

i

G

i

G

i

Gm

Gm

G

i

G

G

G

i

G

G

G

i

G

,,3,213

3

12

2

1

13

3

12

2

11

k

i DiGi PPP iGi PP

n

G

G

m

G

G

n

n

n

m

n

nm

P

P

P

P

P

P

PPP

PPP

1

2

1

1

2

1

2

222

2

1

1

We can use

n

nn

n

q

p

pqPP

PP

P

2

2

2

2

11J

T pq

11J




Penalty Factor Calculation

(covered from book)

• Relationship between power loss and netpower injection

• Use of intermediaries

• Matrix form




Relationship Between Power Loss and NetPower Injection

• No explicit equation for loss.

• Try to relate loss to net power injection.

• From,Where = Generator i power

= Load power at bus i

= Net power injection from bus i

• And, from power flow,

• This means that:

2GP

G1 G2

1GP

3 DP

1 DP

2V

1V

3V

G3

2 DP

3GP

2P

1P

i DiGi PPP

GiP

DiP

iP

n

i

Di

m

i

GiGmGGloss PPPPPP11

32 ,,,

n

i

iGmGGloss PPPPP1

32,,,




Using Intermediaries

• Our goal is to calculate,

• But we know that

• And, net power injection is in terms of

• And, for every bus i,

• We are using as the intermediaries to find

Gi

loss

i

P

P L

1

1

n

iiGmGGloss PPPPP

132 ,,,

Gi

loss

P

P

niP ni ,,2,1,,,32

,θ1 = 0

k

i DiGi PPP




from Net Power Injection

• From net power injection,

and

• We can find,

n

i

iGmGGloss PPPPP1

32,,,

niP ni ,,2,1,,, 32

nk PPPPP

k

n

k

m

k

m

k k

loss ,,3,211

k

lossP




from Chain Rule

• Another way to find

• Using chain rule, we have

• And, from we havei DiGi PPP k

i

k

Gi PP

miP

P

PP

k

Gi

Gi

loss

k

loss ,,3,2

k

lossP

nk P

P

PP

P

PP

k

Gm

Gm

loss

k

G

G

loss

k

loss ,,3,22

2

k

lossP




Two Equations of

• So far, we have two equations representing thesame variable.

• Equating two equations, we have

nk PPPPP

k

n

k

m

k

m

k k

loss ,,3,211

nk P

P

PP

P

PP

k

Gm

Gm

loss

k

G

G

loss

k

loss ,,3,22

2

nk

PPPPPP

P

PP

P

P

k

n

k

m

k

m

k k k

Gm

Gm

loss

k

G

G

loss

,,3,2

1212

2

nk PP

P

PP

P

PPP

k

n

k

m

Gm

loss

k

m

G

loss

k k

,,3,2011 1

2

21

k

lossP




Matrix Form

• Rearranging equation to,

• We can rewrite into following matrix form,

nk PPP

P

PP

P

PP

k k

n

k

Gm

Gm

loss

k

m

G

loss

k

,,3,211 11

2

2

n

Gm

loss

G

loss

n

n

n

m

n

nm

P

P

P

P

P

P

PPP

PPP

1

2

12

2

222

2

1

1

1

1

n

nn

n

q

p

pq

PP

PP

P

2

2

2

2

11J

We can use T pq 11

J




RHS Calculation

• The right hand side of the matrix can be foundfrom,

nk BGV V

Pk k k k k

k ,,3,2cossin 11111

1




Case 3: Solution Outline

• Assume initial dispatch,

Step 1: Solve power flow for

At this step, we can find,

Step 2: Evaluate

and find

Step 3: Check if , DONE.

If not, go to 4.

Step 4: If then,

If then,

Go back to Step 1.

00

3

0

2,,, GmGG PPP

1GP nii ,,3,2

T pq 11

J

niP

i

,,3,21

i

T

pq

ii

PPP

L 111111 J

mi IC L ii ,,2,1

mi IC L ii ,,2,1 GiGi PP

m j IC L j j ,,2,1 GjGj PP




Case 4: With Line Losses,With Generator Limit

• The solution outline to this case is basically thesame as Case 3.

• However, in Step 1 and 4, lower and upper limit

of each generating units need to be check.

Power Control ED

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