Unit Commitment
Economic dispatch gives the optimum schedule corresponding to one particular load on the system. The total load in the power system varies throughout the day and reaches different peak value from one day to another. Different combination of generators, are to be connected in the system to meet the varying load.
When the load increases, the utility has to decide in advance the sequence in which the generator units are to be brought in. Similarly, when the load decrease, the operating engineer need to know in advance the sequence in which the generating units are to be shut down. The problem of finding the order in which the units are to be brought in and the order in which the units are to be shut down over a period of time, say one day, so the total operating cost involved on that day is minimum, is known as Unit Commitment (UC) problem. Thus UC problem is economic dispatch over a day. The period considered may a week, month or a year period.
To solve such problem, simply try all combination of generating units. Some combinations may be infeasible if the sum of all maximum MW for the units committed is less than the load or if the sum of all minimum MW for the units committed is greater than the load. For each feasible combination, units will be dispatched using equal incremental cost rule studied earlier. The results are presented in the Table below. The power system with K generating units (no two identical) must have at least one unit on-line to supply the system load which is never zero over the daily load cycle. If each unit can be considered either on or off, there are 2K - 1 candidate combinations to be examined in each stage of the study period. For example, if K = 4, the 15 theoretically possible combinations for each time interval need to examine. For example, the following are data pertaining to three units in a plant.
|
GU
|
Pmin (MW)
|
Pmax (MW)
|
|
1
|
150
|
600
|
|
2
|
100
|
400
|
|
3
|
50
|
200
|
C1 = 5610 + 79.2 P1 + 0.01562 (P1)2 Rs / h
C2 = 3100 + 78.5 P2 + 0.0194 (P2)2 Rs / h
C3 = 936 + 95.64 P3 + 0.05784 (P3)2 Rs / h
For each feasible combination, units will be dispatched using equal incremental cost rule studied earlier. The results are presented in the Table below.
|
GU-1
|
GU-2
|
GU-3
|
Pmin
|
Pmax
|
P1
|
P2
|
P3
|
C
|
|
off
|
off
|
off
|
|
|
Infeasible
|
|
|
on
|
off
|
off
|
150
|
600
|
550
|
0
|
0
|
53895
|
|
off
|
on
|
off
|
100
|
400
|
Infeasible
|
|
|
off
|
off
|
on
|
50
|
200
|
Infeasible
|
|
|
on
|
on
|
off
|
250
|
1000
|
295
|
255
|
0
|
54712
|
|
off
|
on
|
on
|
150
|
600
|
0
|
400
|
150
|
54188
|
|
on
|
off
|
on
|
200
|
800
|
500
|
0
|
50
|
54978
|
|
on
|
on
|
on
|
300
|
1200
|
267
|
233
|
50
|
56176
|
Note that the least expensive way of meeting the load is not with all the three units running, or any combination involving two units. Rather it is economical to run unit one alone.
Note that to “commit” means a generating unit is to be “turned on”; that is, bring the unit up to speed, synchronize it to the system and make it to deliver power to the network. “Commit enough units and leave them on line” is one solution. However, it is quite expensive to run too many generating units when the load is not large enough. As seen in above example, a great deal of money can be saved by turning units off (decommiting them) when they are not needed.