Unit Committment | Power Systems - Electrical Engineering (EE) PDF Download

 Page 1


11  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 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. This 
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. 
Page 2


11  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 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. This 
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. 
But why is the problem in the operation of electric power system? Why not 
just simply commit enough units to cover the maximum system load and 
leave them running? 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. A great deal of money can be saved by 
turning units off (decommiting them) when they are not needed. 
Example 7 
The following are data pertaining to three units in a plant. The following are data pertaining to three units in a plant. 
Unit 1: Min. = 150 MW; Max. = 600 MW 
  C
1
 = 5610 + 79.2 P
1
 + 0.01562 P
1
2
 Rs / h 
Unit 2: Min. = 100 MW; Max. = 400 MW 
  C
2
 = 3100 + 78.5 P
2
 + 0.0194 P
2
2
 Rs / h 
Unit 3: Min. = 50 MW; Max. = 200 MW 
  C
3
 = 936 + 95.64 P
3
 + 0.05784 P
3
2
 Rs / h 
What unit or combination of units should be used to supply a load of 550 
MW most economically? 
Page 3


11  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 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. This 
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. 
But why is the problem in the operation of electric power system? Why not 
just simply commit enough units to cover the maximum system load and 
leave them running? 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. A great deal of money can be saved by 
turning units off (decommiting them) when they are not needed. 
Example 7 
The following are data pertaining to three units in a plant. The following are data pertaining to three units in a plant. 
Unit 1: Min. = 150 MW; Max. = 600 MW 
  C
1
 = 5610 + 79.2 P
1
 + 0.01562 P
1
2
 Rs / h 
Unit 2: Min. = 100 MW; Max. = 400 MW 
  C
2
 = 3100 + 78.5 P
2
 + 0.0194 P
2
2
 Rs / h 
Unit 3: Min. = 50 MW; Max. = 200 MW 
  C
3
 = 936 + 95.64 P
3
 + 0.05784 P
3
2
 Rs / h 
What unit or combination of units should be used to supply a load of 550 
MW most economically? 
Solution 
To solve this problem, simply try all combination of three units. Some 
combinations will 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. 
Page 4


11  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 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. This 
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. 
But why is the problem in the operation of electric power system? Why not 
just simply commit enough units to cover the maximum system load and 
leave them running? 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. A great deal of money can be saved by 
turning units off (decommiting them) when they are not needed. 
Example 7 
The following are data pertaining to three units in a plant. The following are data pertaining to three units in a plant. 
Unit 1: Min. = 150 MW; Max. = 600 MW 
  C
1
 = 5610 + 79.2 P
1
 + 0.01562 P
1
2
 Rs / h 
Unit 2: Min. = 100 MW; Max. = 400 MW 
  C
2
 = 3100 + 78.5 P
2
 + 0.0194 P
2
2
 Rs / h 
Unit 3: Min. = 50 MW; Max. = 200 MW 
  C
3
 = 936 + 95.64 P
3
 + 0.05784 P
3
2
 Rs / h 
What unit or combination of units should be used to supply a load of 550 
MW most economically? 
Solution 
To solve this problem, simply try all combination of three units. Some 
combinations will 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. 
Unit Min Max 
1 150 600 
2 100 400 
3 50 200 
 
Unit 1 Unit 2 Unit 3 Min. Gen Max. Gen P
1 
P
2 
P
3 
Total cost 
Off Off Off 0 0 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. 
Page 5


11  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 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. This 
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. 
But why is the problem in the operation of electric power system? Why not 
just simply commit enough units to cover the maximum system load and 
leave them running? 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. A great deal of money can be saved by 
turning units off (decommiting them) when they are not needed. 
Example 7 
The following are data pertaining to three units in a plant. The following are data pertaining to three units in a plant. 
Unit 1: Min. = 150 MW; Max. = 600 MW 
  C
1
 = 5610 + 79.2 P
1
 + 0.01562 P
1
2
 Rs / h 
Unit 2: Min. = 100 MW; Max. = 400 MW 
  C
2
 = 3100 + 78.5 P
2
 + 0.0194 P
2
2
 Rs / h 
Unit 3: Min. = 50 MW; Max. = 200 MW 
  C
3
 = 936 + 95.64 P
3
 + 0.05784 P
3
2
 Rs / h 
What unit or combination of units should be used to supply a load of 550 
MW most economically? 
Solution 
To solve this problem, simply try all combination of three units. Some 
combinations will 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. 
Unit Min Max 
1 150 600 
2 100 400 
3 50 200 
 
Unit 1 Unit 2 Unit 3 Min. Gen Max. Gen P
1 
P
2 
P
3 
Total cost 
Off Off Off 0 0 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. 
Example 8 
Daily load curve to be met by a plant having three units is shown below.   
 
 
 
 
 
 
 
 
 
1200 
MW 
 
 
 
 
 
 
 
 
Data pertaining to the three units are the same in previous example. 
Starting from the load of 1200 MW, taking steps of 50 MW find the shut-
down rule. 
500 
MW 
12 noon 
6 am 2 am 8 pm 4 pm 
12 noon