Page 1
9 OPTIMUM SCHEDULING OF SYSTEM LOAD BETWEEN PLANTS
- SOLUTION PROCEDURE
To determine the optimum scheduling of system load between plants, the
data required are i) system load, ii) incremental cost characteristics of the
plants and iii) loss coefficient matrix. The iterative solution procedure is:
Step 1
For the first iteration, choose suitable initial value of ?. While finding this,
one way is to assume that the transmission losses are zero and the plants
are loaded equally.
Step 2
Substitute the value of ? into the coordination equations
?
P
P
?
P d
C d
i
L
i
i
=
?
?
+ i = 1,2,………..,N
i.e. ? P B 2 ? ) ß P a 2 (
n
N
1 n
mn i i i
= + +
?
=
i = 1,2,………..,N
The above set of linear simultaneous equations are to be solved for the
values s P
'
i
by some efficient means. Gauss Seidel method is
recommended.
Page 2
9 OPTIMUM SCHEDULING OF SYSTEM LOAD BETWEEN PLANTS
- SOLUTION PROCEDURE
To determine the optimum scheduling of system load between plants, the
data required are i) system load, ii) incremental cost characteristics of the
plants and iii) loss coefficient matrix. The iterative solution procedure is:
Step 1
For the first iteration, choose suitable initial value of ?. While finding this,
one way is to assume that the transmission losses are zero and the plants
are loaded equally.
Step 2
Substitute the value of ? into the coordination equations
?
P
P
?
P d
C d
i
L
i
i
=
?
?
+ i = 1,2,………..,N
i.e. ? P B 2 ? ) ß P a 2 (
n
N
1 n
mn i i i
= + +
?
=
i = 1,2,………..,N
The above set of linear simultaneous equations are to be solved for the
values s P
'
i
by some efficient means. Gauss Seidel method is
recommended.
Step 3
Compute the transmission loss P
L
from P
L
= [ P ] [ B ] [ P
t
]
where [ P ] = [
N 2 1
.P .......... P P ] and [ B ] is the loss coefficient matrix.
Step 4
Compare
?
=
N
1 i
i
P with P
D
+ P
L
to check the power balance. If the power
balance is satisfied within a specified tolerance, then the present solution
is the optimal solution; otherwise update the value of ?. is the optimal solution; otherwise update the value of ?.
First time updating can be done judiciously. Value of ? is increased or
decreased by about 5% depending on
?
=
N
1 i
i
P < < < < P
D
+ P
L
or
?
=
N
1 i
i
P > > > > P
D
+ P
L
In the subsequent iterations, using linear interpolation, value of ? can be
updated as
Page 3
9 OPTIMUM SCHEDULING OF SYSTEM LOAD BETWEEN PLANTS
- SOLUTION PROCEDURE
To determine the optimum scheduling of system load between plants, the
data required are i) system load, ii) incremental cost characteristics of the
plants and iii) loss coefficient matrix. The iterative solution procedure is:
Step 1
For the first iteration, choose suitable initial value of ?. While finding this,
one way is to assume that the transmission losses are zero and the plants
are loaded equally.
Step 2
Substitute the value of ? into the coordination equations
?
P
P
?
P d
C d
i
L
i
i
=
?
?
+ i = 1,2,………..,N
i.e. ? P B 2 ? ) ß P a 2 (
n
N
1 n
mn i i i
= + +
?
=
i = 1,2,………..,N
The above set of linear simultaneous equations are to be solved for the
values s P
'
i
by some efficient means. Gauss Seidel method is
recommended.
Step 3
Compute the transmission loss P
L
from P
L
= [ P ] [ B ] [ P
t
]
where [ P ] = [
N 2 1
.P .......... P P ] and [ B ] is the loss coefficient matrix.
Step 4
Compare
?
=
N
1 i
i
P with P
D
+ P
L
to check the power balance. If the power
balance is satisfied within a specified tolerance, then the present solution
is the optimal solution; otherwise update the value of ?. is the optimal solution; otherwise update the value of ?.
First time updating can be done judiciously. Value of ? is increased or
decreased by about 5% depending on
?
=
N
1 i
i
P < < < < P
D
+ P
L
or
?
=
N
1 i
i
P > > > > P
D
+ P
L
In the subsequent iterations, using linear interpolation, value of ? can be
updated as
?
? ?
= -
-
+
- +
-
-
+ =
N
1 i
k
i
k
L D
N N
1 k
i
k
i
1 k k
k 1 k
] P P P [
P P
? ?
? ? (41)
?
k
i
P
?
-1 k
i
P
1 k
?
+
1 k
?
-
k
?
P
D
+ P
L
k
? ?
= =
-
1 i 1 i
i i
P P
Here k-1, k and k+1 are the previous iteration count, present iterative
count and the next iteration count respectively.
Step 5
Return to Step 2 and continue the calculations of Steps 2, 3 and 4 until the
power balance equation is satisfied with desired accuracy.
The above procedure is now illustrated through an example.
Page 4
9 OPTIMUM SCHEDULING OF SYSTEM LOAD BETWEEN PLANTS
- SOLUTION PROCEDURE
To determine the optimum scheduling of system load between plants, the
data required are i) system load, ii) incremental cost characteristics of the
plants and iii) loss coefficient matrix. The iterative solution procedure is:
Step 1
For the first iteration, choose suitable initial value of ?. While finding this,
one way is to assume that the transmission losses are zero and the plants
are loaded equally.
Step 2
Substitute the value of ? into the coordination equations
?
P
P
?
P d
C d
i
L
i
i
=
?
?
+ i = 1,2,………..,N
i.e. ? P B 2 ? ) ß P a 2 (
n
N
1 n
mn i i i
= + +
?
=
i = 1,2,………..,N
The above set of linear simultaneous equations are to be solved for the
values s P
'
i
by some efficient means. Gauss Seidel method is
recommended.
Step 3
Compute the transmission loss P
L
from P
L
= [ P ] [ B ] [ P
t
]
where [ P ] = [
N 2 1
.P .......... P P ] and [ B ] is the loss coefficient matrix.
Step 4
Compare
?
=
N
1 i
i
P with P
D
+ P
L
to check the power balance. If the power
balance is satisfied within a specified tolerance, then the present solution
is the optimal solution; otherwise update the value of ?. is the optimal solution; otherwise update the value of ?.
First time updating can be done judiciously. Value of ? is increased or
decreased by about 5% depending on
?
=
N
1 i
i
P < < < < P
D
+ P
L
or
?
=
N
1 i
i
P > > > > P
D
+ P
L
In the subsequent iterations, using linear interpolation, value of ? can be
updated as
?
? ?
= -
-
+
- +
-
-
+ =
N
1 i
k
i
k
L D
N N
1 k
i
k
i
1 k k
k 1 k
] P P P [
P P
? ?
? ? (41)
?
k
i
P
?
-1 k
i
P
1 k
?
+
1 k
?
-
k
?
P
D
+ P
L
k
? ?
= =
-
1 i 1 i
i i
P P
Here k-1, k and k+1 are the previous iteration count, present iterative
count and the next iteration count respectively.
Step 5
Return to Step 2 and continue the calculations of Steps 2, 3 and 4 until the
power balance equation is satisfied with desired accuracy.
The above procedure is now illustrated through an example.
EXAMPLE 4
Consider a power system with two plants having incremental cost as
MWh / Rs 200 P 1.0 IC
1 1
+ = MWh / Rs 150 P 1.0 IC
2 2
+ =
Loss coefficient matrix is given by B =
?
?
?
?
?
?
-
-
0.0024 0.0005
0.0005 0.001
Find the optimum scheduling for a system load of 100 MW.
SOLUTION
Assume that there is no transmission loss and the plants are loaded Assume that there is no transmission loss and the plants are loaded
equally. Then MW 50 P
1
= . Initial value of ? = ( 1.0 x 50 ) + 200 = 250 Rs /
MWh.
Coordination equations:
250 ) P 0.0048 P 0.001 ( 250 150 P 1.0
250 ) P 0.001 P 0.002 ( 250 200 P 1.0
2 1 2
2 1 1
= + - + +
= - + +
i.e. 50 P 0.25 P 1.5
2 1
= - i.e.
2 1
P 0.1667 33.3333 P + =
100 P 2.2 P 0.25
2 1
= + -
1 2
P 0.1136 45.4545 P + =
These equations are solved by Gauss Seidel method.
Page 5
9 OPTIMUM SCHEDULING OF SYSTEM LOAD BETWEEN PLANTS
- SOLUTION PROCEDURE
To determine the optimum scheduling of system load between plants, the
data required are i) system load, ii) incremental cost characteristics of the
plants and iii) loss coefficient matrix. The iterative solution procedure is:
Step 1
For the first iteration, choose suitable initial value of ?. While finding this,
one way is to assume that the transmission losses are zero and the plants
are loaded equally.
Step 2
Substitute the value of ? into the coordination equations
?
P
P
?
P d
C d
i
L
i
i
=
?
?
+ i = 1,2,………..,N
i.e. ? P B 2 ? ) ß P a 2 (
n
N
1 n
mn i i i
= + +
?
=
i = 1,2,………..,N
The above set of linear simultaneous equations are to be solved for the
values s P
'
i
by some efficient means. Gauss Seidel method is
recommended.
Step 3
Compute the transmission loss P
L
from P
L
= [ P ] [ B ] [ P
t
]
where [ P ] = [
N 2 1
.P .......... P P ] and [ B ] is the loss coefficient matrix.
Step 4
Compare
?
=
N
1 i
i
P with P
D
+ P
L
to check the power balance. If the power
balance is satisfied within a specified tolerance, then the present solution
is the optimal solution; otherwise update the value of ?. is the optimal solution; otherwise update the value of ?.
First time updating can be done judiciously. Value of ? is increased or
decreased by about 5% depending on
?
=
N
1 i
i
P < < < < P
D
+ P
L
or
?
=
N
1 i
i
P > > > > P
D
+ P
L
In the subsequent iterations, using linear interpolation, value of ? can be
updated as
?
? ?
= -
-
+
- +
-
-
+ =
N
1 i
k
i
k
L D
N N
1 k
i
k
i
1 k k
k 1 k
] P P P [
P P
? ?
? ? (41)
?
k
i
P
?
-1 k
i
P
1 k
?
+
1 k
?
-
k
?
P
D
+ P
L
k
? ?
= =
-
1 i 1 i
i i
P P
Here k-1, k and k+1 are the previous iteration count, present iterative
count and the next iteration count respectively.
Step 5
Return to Step 2 and continue the calculations of Steps 2, 3 and 4 until the
power balance equation is satisfied with desired accuracy.
The above procedure is now illustrated through an example.
EXAMPLE 4
Consider a power system with two plants having incremental cost as
MWh / Rs 200 P 1.0 IC
1 1
+ = MWh / Rs 150 P 1.0 IC
2 2
+ =
Loss coefficient matrix is given by B =
?
?
?
?
?
?
-
-
0.0024 0.0005
0.0005 0.001
Find the optimum scheduling for a system load of 100 MW.
SOLUTION
Assume that there is no transmission loss and the plants are loaded Assume that there is no transmission loss and the plants are loaded
equally. Then MW 50 P
1
= . Initial value of ? = ( 1.0 x 50 ) + 200 = 250 Rs /
MWh.
Coordination equations:
250 ) P 0.0048 P 0.001 ( 250 150 P 1.0
250 ) P 0.001 P 0.002 ( 250 200 P 1.0
2 1 2
2 1 1
= + - + +
= - + +
i.e. 50 P 0.25 P 1.5
2 1
= - i.e.
2 1
P 0.1667 33.3333 P + =
100 P 2.2 P 0.25
2 1
= + -
1 2
P 0.1136 45.4545 P + =
These equations are solved by Gauss Seidel method.
2 1
P 0.1667 33.3333 P + =
1 2
P 0.1136 45.4545 P + =
Iteration count
1
P
2
P
0 0 0
1 33.3333 49.2411
2 41.5418 50.1736
3 41.6972 50.1913
4 41.7002 50.1916
[ ]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
50.1916
41.7002
0.0024 0.0005
0.0005 0.001
50.1916 41.7002 P
L
= [ ]
?
?
?
?
?
?
0.09961
0.01660
50.1916 41.7002 = 5.6918 MW
MW 91.8918 P P
2 1
= + MW 105.6918 P P
L D
= +
Since
2 1
P P + <
L D
P P + , ? value should be increased. It is increased by 4 %.
New value of ? = 250 x 1.04 = 260 Rs / MWh.
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