Post on 06-Apr-2018
8/3/2019 Economic Dispatch Termal 1
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ECONOMIC DISPATCHUNIT-UNIT PEMBANGKIT
TERMAL
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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
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PN
P2
P1
FN
Pload
F2
F1
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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.
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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
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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
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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
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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
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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
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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
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λ (3)
Error :
(∑ Pi)-PR
λ
(3)
(2)
(1)
λ (2)
λ (1)
Lambda Projections
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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 :
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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
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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
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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)
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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 :
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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
<