economic dispatch termal 1

8/3/2019 Economic Dispatch Termal 1

http://slidepdf.com/reader/full/economic-dispatch-termal-1 1/18

ECONOMIC DISPATCHUNIT-UNIT PEMBANGKIT

TERMAL



INTRODUCTION

• Introduce techniques of power system

optimization

• Introduce Lagrange Multiplier method of

solution

• The system consists of N thermal generating

units connected to a single bus-bar servinga received electrical load Pload



PN

P2

P1

FN

Pload

F2

F1



The input to each unit, shown as F i , represents

the cost rate of the unit.

The output of each unit, Pi , is the electrical

power generated by that particular unit.

The total cost rate of this system is, the sum of

the costs of each of the individual unit.

The essential constraint on the operation of thissystem is that the sum of the output powers must

equal the load demand.



Problem Formulation :

N

i

ii

N T

PF

F F F F F

1

321

)(

...........

The objective function, F T ,, is equally to the total cost

for supplying the indicated load.

The problem is to minimize F T subject to the constraint

that the sum of the powers generated must equal the

received load.

Transmission losses are neglected

N

i

iload PP

1

0



LAGRANGE function :

L T F

The necessary conditions for an extreme value of the

objective function :

0d

)(d

i

ii

iP

PF

P

L

i

i

P

F

d

d0

, atau



The necessary conditions and inequalities for the existence

of a minimum cost-operating condition :

i

i

P

F

d

d

max,min, iiiPPP

load

N

ii

PP

1

N equations

2N inequalities

1 constraint



i

i

P

F

d

d

i

i

P

F

d

d

i

i

P

F

d

d

max,iiPP

max,min, iiiPPP

min,ii

PP

for

for

for



Economic Dispatch by

Lambda Iteration method

END

PRINT SCHEDULE

TOLERANCE?

FIRST ITERATION?

CALCULATE

CALCULATE Pi

FOR i = 1 ….. N

SET λ

START

N

i

iload PP

1

PROJECT λ

YES

NO

NO

YES



Graphical solution to economic dispatch :

P2(MW)

2

2

dP

dF

(R/MWh)

P1(MW)

1

1

dP

dF

(R/MWh)

P3(MW)

3

3

dP

dF

(R/MWh)

++

+

PR

= P1

+ P2

+ P3



λ (3)

Error :

(∑ Pi)-PR

λ

(3)

(2)

(1)

λ (2)

λ (1)

Lambda Projections



EXAMPLE 3A :

2

11100142.02.70.510 PP

h

MBtu H

2

222 00194.085.70.310 PPh

MBtu H

2

333 00482.097.70.78 PPh

MBtu H

No.Unit SystemInput Input-Output Curve Output-Max

(MW)

Output-Min

(MW)

Fuel Cost(R/Mbtu)

Unit 1 Coal-Fired

Steam

600 150 1.1

Unit 2 Oil-Fired

Steam

400 100 1.0

Unit 3 Oil-Fired

Steam

200 50 1.0

2

111 00142.02.70.510 PPh

MBtu H

2

222 00194.085.70.310 PPh

MBtu H

2

333 00482.097.70.78 PPh

MBtu H

F1(P1) = H1(P1) x 1.1 = 561 + 7.92 P1 + 0.001562P1

2

R/hF2(P2) = H2(P2) x 1.0 = 310 + 7.85 P2 + 0.00194P2

2 R/h

F3(P3) = H3(P3) x 1.0 = 78 + 7.97 P3 + 0.00482P32 R/h

Karakteristik Input-Output dalam R/h :



1

1

1 003124.092.7 PdP

dF

2

2

2 00388.085.7 P

dP

dF

3

3

3 00964.097.7 PdP

dF

P1 + P2 + P3 = 850 MW

= 9.148 R/MWh

P1 = 393.2 MW

P2 = 334.6 MW

P3 = 122.2 MW

Optimal Economic

Solution



EXAMPLE 3B :

No.Unit SystemInput Input-Output Curve Output-Max

(MW)

Input-Min

(MW)

Fuel Cost(R/Mbtu)

Unit 1 Coal-Fired

Steam

600 150 0.9

Unit 2 Oil-Fired

Steam

400 100 1.0

Unit 3 Oil-Fired

Steam

200 50 1.0

2

111 00142.02.70.510 PPh

MBtu H

2

222 00194.085.70.310 PPh

MBtu H

2

333 00482.097.70.78 PPh

MBtu H

Karakteristik Input-Output dalam R/h :

F1(P1) = H1(P1) x 0.9 = 459 + 6.48 P1 + 0.00128P12 R/h

F2(P2) = H2(P2) x 1.0 = 310 + 7.85 P2 + 0.00194P22 R/h

F3(P3) = H3(P3) x 1.0 = 78 + 7.97 P3 + 0.00482P32 R/h



1

1

1 00256.048.6 PdP

dF

2

2

2 00388.085.7 P

dP

dF

3

3

3 00964.097.7 PdP

dF

P1 + P2 + P3 = 850 MW

= 8.284 R/MWh

P1 = 704.6 MW

P2 = 111.8 MW

P3 = 32.6 MW

P1 > 600 MW (max)

P3 < 50 MW (min)



Unit 1 diset output maximum dan unit 3 output minimum

P1 = 600 MWP2 = 200 MW

P3 = 50 MW

MWh RdP

dF

P

/ 626.82002

2

2

MWh RdP

dF

P

/ 016.86001

1

2

MWh RdP

dF

P

/ 452.8

503

3

3

Peningkatan biaya pada unit 2 :

Peningkatan biaya pada unit 1dan 3 :



Peningkatan biaya pada unit 1 paling kecil di MAX -kan

P1 = 600 MW

2

2

2 00388.085.7 PdP

dF

3

3

3 00964.097.7 PdP

dF

P2 + P3 = 850 – P3 = 250 MW

= 8.576 R/MWh

P2 = 187.1 MW

P3 = 62.9 MW

MWh RdP

dF

MW P

/ 016.8

6001

1

1

3

3

2

2

dP

dF dan

dP

dF

<



N THERMAL UNITSSERVING LOAD

THROUGHTRANSMISSION NETWORK

NETWORK LOSSES CONSIDERED

economic dispatch termal 1

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