EE4031 2 Economic Operation

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The HK Polytechnic University Economic Operation

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1 Economic Load Dispatch

There are many factors involved in the successful operation of a power system.

Next to security, economy is the most important.

Economic load dispatch concerns with the optimum load scheduling of the

different generating plants in the system and it involves 2 different problems.

1. Economic Dispatch an on-line problem to optimally distribute the load

among the generating units to minimise the total cost of supply while the total

demand and the losses at any instant is met by the total generation.

2. Unit Commitment a predispatch problem to select optimally out of the

available generating units to meet the expected load and provide a specified

margin of operating reserve over a specified period of time.

Economic Load Dispatch only deals with the problem of minimum cost of power

production. Other economic operation aspects such as minimum-loss delivery

of the generated power to the loads are beyond the scope of this topic.

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2 Economic Dispatch

The sole focus of economic dispatch is the minimisation of the thermal fuel costs,

even though there are other factors such as labour, supplies and maintenance in

dealing with the total costs of producing electrical energy.

There are 2 important cases of optimal economic dispatch to be analysed.

1. Economic Dispatch Neglecting Losses

(a) Generators in a given station there is no transmission losses within the

power station.

(b) Urban systems small power systems with short transmission lines where

the line losses are negligible.

2. Economic Dispatch Including Transmission Losses large electric power

systems have many widely separated generating units which are interconnected

by long transmission line, it is necessary to consider the associated losses in

determining system optimal economic dispatch.

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2.1 Generator Operating Cost

The operating cost of a thermal plant is mainly the cost of the fuel. The fuel is often

coal, oil, gas or nuclear. Other costs such as costs of maintenance, labour and

supplies contribute only to a small extent and are assumed to vary as a fixed

percentage of the fuel cost.

The fuel cost is meaningful in case of thermal stations, but for hydro stations where

the energy storage is apparently free, the operating fuel cost as such is not

meaningful. This is the reason why only thermal plants are considered in the

following sections.

2.2 Fuel-Cost Characteristics and Incremental Fuel Cost

The cost-output curves of generating units of a thermal plant are important to

describe the efficiency of the plant. A typical cost-output curve is shown below.

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In practice, the operating cost Ci is constructed of

piecewise non-linear functions valid for ranges of output

Pibased on empirical data. By fitting a suitable degree

polynomial, an analytical expression for operating cost

can be written asCi(Pi). It generally suffices to fit a

second degree polynomial, i.e.

Ci =B2P2i + B1Pi+ B0 $/h

whereB0,B1 andB2 are constants depending upon

a particular plant and Pi,min Pi Pi,max.

The slope of the cost curve, i.e. dCidPi, is called the

incremental fuel cost (ICi). If the cost is approximated

as a quadratic as in the above, then ICi is linear.

ICi= 2B

2Pi+ B

1 $/MWh

Unitoperatingcost,Ci

$/h

Unit output, PiMW

dCi

dPi

Cost-Output Curve

Incrementalcost,dCi/dPi

Unit output, PiMW

Incremental Cost Curve

$/MWh

Linear

Approximation

Pi,min Pi,max

Polynomial

Approximation

Piecewise

Non-linear

Characteristics

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2.3 Optimum Loading of Two Units Within a Plant

Consider 2 generators either in the same plant or close enough electrically

so that the line losses may be neglected. Let

C1 = fuel cost of generator 1 for a power output of P1

C2 = fuel cost of generator 2 for a power output of P2

C = total fuel cost = C1+ C2

PD= total load demand = P1+ P2 = constant

Then CP1= C1P1 +

C2P1

= C1P1

+ C2P2

P2P1

and PDP1= 1+ P2P1 = 0

P2P1

= 1

For minimum cost, CP1= 0 C1P1 =

C2P2

= (say)

i.e. the basic criterion for economical division of load between units is that

both the units must operate at the same incremental fuel cost .

This is known as the equal criterion.

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2.4 Example 1

The total amount output of a 2-generator station is PMW and the cost curves of

the generators are: C1=B2P21 + B1P1+ B0 $/h whereP =P1+ P2

C2=B2P

22 + B

1P2+ B

0 $/h

Determine how this load should be shared to give the most economical distribution.

The incremental costs of the generators are: IC1= 2B2P1+ B1

IC2= 2B2P2+ B1

For optimum division of load, the two incremental costs should be equal, i.e.

2B2P1+ B1= 2B2P2+ B

1 (E1.1)

Eliminate P2 and rearrange (E1.1): P1= B2

B2+ B2P+

B1 B12(B2+ B2)

Eliminate P1 and rearrange (E1.1): P2= B2

B2

+ B2

P+ B1 B

1

2(B2

+ B2

)

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2.5 Example 2

Determine the most economical load distribution and the total minimum costof operation at this load as described in Example 1. Given that:

P= 450MW B1 = 4.0 $/MWh B1= 3.4 $/MWh

B2 = 0.02 $/MW2h B2= 0.01$/MW

2h

From the results obtained in Example 1, the optimium load of generator 1 and 2 are:

P1= B2

B2+ B2P+

B1 B12(B2+ B2)

=

0.01

0.02 + 0.01 450 +

3.4 4.0

2(0.02 + 0.01) = 140MWP2=P P1= 450 140 = 310MW

The total minimum cost of operation:

C= C1+ C2 =B2P21 + B1P1+ B0+ B

2P

22 + B

1P2+ B

0

= 0.02(140)2 + 4(140) + B0+ 0.01(310)2 + 3.4(310) + B0

= 2967 + B0+ B0 $/h

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2.6 Optimum Loading of N Units Within a Plant

Consider n generating units within a plant. Let C1, C2 . . . C n be the fuel costs of

individual units for the corresponding power outputs P1, P2 . . . P n respectively. If

C is the total fuel cost of the entire system and PD is the total load demand.

C =C1+ C2+ . . .+ Cn=n

i=1 Ci

PD=P1 + P2 + . . .+ Pn = ni=1 Pi=constant

For a minimum C, the total differential dC= 0, i.e.

dC= C

P1dP1+

C

P2dP2+ . . . +

C

PndPn = 0 (1)

For the total load demand PD to remain constant, dPD = 0, i.e.

dPD =dP1+ dP2+ . . . + dPn= 0 (2)

Multiplying (2) by and then subtracting from (1) gives

CP1

dP1

+ CP2

dP2

+ . . .+ CPn

dPn

= 0 (3)

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Equation (3) is satisfied if each term is equal to zero, i.e.

CP1

= CP2

=. . .= CPn

=

OrdC

dP1=

dC

dP2=. . .=

dC

dPn= coordination equations

as Ci is a function of Pi only, i.e.CiPi

= dCidPi

This shows that the criterion for most economical division of load between units

within a plant is that all the units must operate at the same incremental fuel cost .

This is known as the principle of equal criterion for economic operation.

The method used above is called the method of Lagrangian multipliers.

The Lagrangian is defined as: L=n

i=1 Ci (n

i=1 Pi PD)

Minimization is achieved by the condition:L

Pi= 0

dCidPi

=

10

2.7 Example 3

The power needs of a large plant are served by three generating unit, which have

the following incremental cost functions:

IC1= 2B2P1 + B1= 0.010P1+ 8.8 $/MWh

IC2= 2B2P2 + B

1= 0.015P2+ 10.2 $/MWh

IC3= 2B2 P3+ B

1= 0.020P3+ 12.1 $/MWh

Determine the optimal economic dispatch for a total power demand ofPD = 800MW and the cost in cents per kWh at this operating condition.

The required optimal condition is when = IC1 = IC2 = IC3 hence

P1= 8.80.01 = 100( 8.8)

P2= 10.20.015

= 66.67( 10.2)

P3= 12.1

0.02 = 50( 12.1)

Next, substitute the above equations into 3

i=1Pi = PD yields

100( 8.8) + 66.67( 10.2) + 50( 12.1) = 800

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Hence, = 13.6844 $/MWh = 1.36844cents/kWh

P1= 488.44 MW P2 = 232.31MW P3= 79.22MW

2.8 Computer Solution for Optimum Loading of Generators

It is not uncommon for the solution in the lossless case to be determined easily.

However, analytical solutions are often difficult to obtain when losses in the

transmission network must be considered and efforts are better served by resorting

to an iteration solution using computer as follows:

(a) Express each generator power in terms of the Lagrangian multiplier (b) Pick a suitable initial value of

(c) Find the corresponding real powers P1, P2 . . . P n

(d) If

Pi PD = 0, the optimal solution is reached

If

Pi PD 0, decrease and goto (c)

12

+

dC1

dP1

Compared to Pd

dC2

dP2

dC3

dP3

++

P3P2P1

dC1

dP1

dC2

dP2

dC3

dP3

P3P2P1

outside limits of Gnerator 1 & 3

dCi

dPi

P

G1

G2

Range 3

Range 2

Range 1

Graphical Representation

Solving each generator outputusing successive adjustments

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2.9 Example 4

The following are the fuel-cost curves for the 3 generating units of a power plant.C1 = 0.0015P

2

1 + 8.0P1+ 300 $/h 50 P1 400MW

C2 = 0.0005P2

2 + 8.0P1+ 450 $/h 50 P2 800MW

C3 = 0.0010P2

3 + 7.5P1+ 700 $/h 50 P3 1000MW

Plot the graph of incremental cost versus total load (150 MW PD 2200 MW)

when the generators are sharing the load optimally, and hence find the incremental

cost for the load PD=1800 MW.

Incremental Cost ($/MWh) Generator Total Power,PD (MW)

IC1 = 0.003P1+ 8.0 7.6 G3 50+50+50 = 150

IC2 = 0.001P2+ 8.0 8.05 G2,G3 50+50+275 = 375

IC3 = 0.002P3+ 7.5 8.15 G1,G2,G3 50+150+325 = 525

8.8 G1,G3 266.67+800+650 = 1716.67

9.2 G3 400+800+850 = 2050

9.5 400+800+1000 = 2200

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3 Economic Dispatch Including Transmission Losses

When transmission losses are included in the economic dispatch problem, the totalload demand equation becomes:

ni=1

Pi PL = PD

where PL is the total transmission loss. In general, PL is not constant but depends

on the unit outputs Pi. A new Lagrangian function L is generated such that

L= C n

i=1

Pi PL PDMinimum of Cis attained when LPi = 0for all values of i, i.e.

C

Pi

Pi

ni=1

Pi PL PD

= 0

or dCi

dPi

1

PL

Pi

= 0 (4)

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or dCi

dPi

1

1 PLPi

=

or dCi

dPiLi = for all values of i

where Li = 1

1 PLPiis known as the penalty factor for plant i.

Thus the optimum fuel economy is achieved when the product of the incremental

fuel times the penalty factor is the same for all plants. PLPi

is known as the

incremental transmission loss at plant i and is known as the incremental cost of

received power in $/MWh.

dC1

dP1L1 =

dC2

dP2L2 = . . .=

dCn

dPnLn =

The above equations are known as the exact coordination equations because both

incremental fuel cost and incremental transmission loss are coordinated for best

economy.

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3.1 Loss Penalty Factors

Focusing on the generator i and assuming that it undergoess an increase ingenerating power of amountPi, which in turn manifests itself as an increased

share of total load demand as well as associated line losses. That is:

Pi= PD+ PL

The loss penalty factor for this generator can be rewritten as:

Li = 1

1 PLPi

1

1 PLPi=

PiPi PL

= PiPD

1

or dCidPiLi = dCidPi

PiPD= dCidPD

= incremental cost ofreceived power for unit i

Li 1: generator is close to the loads and most of the increased generator

power goes to supply the load demand.

Li>1 : generator is far away from loads and is required to supply an

increased load demand in the presence of high line losses.

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3.2 Loss Formula Concept B Coefficients

The two generator units of this system are assumed to deliver power to a remotely

located load with power factorpf1 andpf2. Three transmission line segments are

involved with resistances of R1, R2 and R3.

Let the currents supplied by G1and G2

to the load are I1 and I2 and are co-

phasal, the line losses PL is:

V1

pf1

R1

I1

R2

I2

R3

Load

V2

pf2

G1 G2

PL= |I1|2R1+ |I2|

2R2+ (|I1|+ |I2|)2R3

=

R1+ R3V21(pf1)

2

P21 +

2R3

V1V2(pf1)(pf2)

P1P2+

R2+ R3V22(pf2)

2

P22

=B11P21 + 2B12P1P2+ B22P

22

The B coefficients can be assumed constant if bus voltages and power factors

remain constant. In practice, they are calculated for some average operating

condition and can be used reliably for a faily wide range around those conditions.

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3.3 Loss Penalty Factors Via Loss Coefficients

Based on the experience with the simple case in last section, a general expression

for the total system losses can be written, with several approximations, as:

PL=ni=1

nj=1

PiBijPj andPLPj

=ni=1

2BijPj

where Bij is the loss coefficients which are constants unders certain assumed

operating conditions and Bij =Bji . Assuming quadratic plant cost curves as:

Ci=Bi2P

2i + B

i1Pi+ B

i0

The incremental cost is: dCi

dPi= 2Bi

2Pi+ Bi

1

Substituting PLPjand dCidPi

from the above coordination equation (4) an collecting all

terms of Pi:Pi =

1Bi

1

j=i2BijPj2Bi

2

+ 2Bii

For any particular value of , Pi can be solved iteratively by assuming initial values

of P i. A convenient choice is Pi= 0.

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3.4 Example 5

Generator 1 is remotely located and generator 2 is at the load centre.

When 100 MW is dispatched from generator 1, the transmission loss is 10%.

When the system is optimally dispatched at a certain time the system

is found to be $25 per MWh.

Determine the power output of each generator and the power taken by the load.

The following cost data is supplied.

Generator 1: 0.015P2 + 17P+ 14 $/h

Generator 2: 0.030P2 + 19P+ 16 $/h

Solution: PL =B11P21

B11 = PLP21

= 10010%1002 = 103 MW1

Incremental cost : IC1= 0.03P1+ 17

IC2= 0.06P2+ 19

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Exact coordination equation : IC1L1 =I C2L2=

where L1= ?L2= 1 (no loss)

= 25 (given)

Hence, 0.06P2+ 19 = 25

P2= 100 MW

Penalty factor : L1= 1

1 PLP1=

1

1 2B11P1

(0.03P1+ 17)( 112B11P1 ) = 25

P1= 8

0.03 + 50B11=

8

0.08

= 100 MW

Load, PD=

Pi PL= 100 + 100 10 = 190 MW

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3.5 Example 6

Consider the operation of 3 thermal generators G1, G2 and G3 in a small system.

The rating of the units are G1 = 150MW and G2 = G3 = 350MW. The operating

costs are:

C1 = 0.25P2G1+ 60PG1+ 15 $/h

C2 = 0.25P2G2+ 70PG2+ 13 $/h

C3 = 0.25P2

G3+ 75PG3+ 12 $/hThe transmission losses are given by a loss formula whose only non-zero coefficient

is B33 = 0.0005 MW1.

Assuming that G1s power output is fixed at 140MW, calculate to an accuracy of

2MW per generator the optimal dispatch of a system demand of 550MW and

obtain the corresponding average system electricity production cost in cents/kWh.

Determine also the marginal cost of supplying an additional kWh of electricity to the

system under these operating conditions.

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Solution: IC1 = 0.5P1+ 60 PD = 550 MW= PiIC2 = 0.5P2+ 70 PL =B33P

23

IC3 = 0.5P3+ 75 PL

P3= 2B33P3 Li =

1

1PLPi

P1= 140 MW , L1 =L2= 1 & L3= ?

1st iteration : set L3= 1 PL= 0 (lossless)

P2+ P3=PD P1+ PL = 410 MW

IC2L2=I C3L3 0.5P2+ 70 = 0.5P3+ 75

P2= 210 MW

P3= 200 MW

2nd iteration : L3 = 1

12(0.0005)(200)= 1.25

PL= (0.0005)(2002) = 20

P2+ P3= 550 140 + 20 = 430 MW

0.5P2+ 70 = 1.25(0.5P3+ 75)

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P2= 260 MW

P3= 170 MW

3rd iteration : L3 = 1

12(0.0005)(170) = 1.2048

PL= (0.0005)(1702) = 14.45

P2+ P3= 424.45 MW

0.5P2+ 70 = 1.2048(0.5P3+ 75)

P2= 250.4 MWP3= 174 MW

4th iteration : L3 = 1.2107 P2 = 251.82 MW

PL = 15.14 MW P3 = 173.63 MW

Hourly generating cost = (C1+ C2+ C3)/550

= 122.5 $/MWh= 12.25 cents/kWh

Marginal cost per kWh =I C2L2 =I C2= 70 + 0.5(251.82)

= 195.9 $/MWh= 19.59cents/kWh

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4 Interconnection for Interchange of Power

Power plants are connected together through tie-line (transmission line) to form alarge interconnected system due to the following advantages.

1. Increased Reliability - extra spining reserve and redundant power paths

The loss of generator or transmission line can be made up from spinning

reserve among generators throughout the interconnection or alternative

transmission paths.

2. Cost Reduction - power interchange

Power is bought from the lower cost generator and hence the cost of

generation is reduced.

Regional spinning reserve is reduced by relying on system reserve, thus

saving operation cost.

3. Improved Regulation

Changes of load can be catered by all generating units within the whole

system and not the generators in the region only.

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4.1 Interchange Evaluation

Consider two power systems operating with different incremental costs (IC).

Utility A is generating at a lower IC than utility B.

If utility B buys power from utility A for its load at a price less than the cost of

generation by itself, utility B would save money.

Utility A would benefit economically from selling power to utility B as long as

utility B is willing to pay a price greater than utility As cost of generation.

Steps for interchange evaluation:

1. Assume no power interchange, calculate the economis dispatch (ED) for each

system alone.

2. Determine which system has the lower IC. The system with lower IC runs a

series of ED, each having a greater demand. The system with higher IC runs a

series of ED, each having a lower demand.

3. Find out at which level of interchange energy will bring the two systems towards

the most economic operation.

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4.2 Example 7

Consider the following interconnected areas:

Area 1

Load 700MW

5

4

2

6

1

3

Area 2

Load 1100MW

Ci =fi(ai+ biPi+ ciP2i) P

mini Pi P

maxi

Unit fi ai bi ci Pmini (MW) P

maxi (MW)

1 2.0 561 7.92 0.001562 150 600

2 2.0 310 7.85 0.00194 100 400

3 2.0 78 7.97 0.00482 50 200

4 1.9 500 7.06 0.00139 140 590

5 1.9 295 7.46 0.00184 110 440

6 1.9 295 7.46 0.00184 110 440

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Step 1: Perform separate economic dispatch for each area

Area 1 Area 2

P1 = 322.7MW, P2 = 277.9MW P4 = 524.7MW, P5 = 287.7MW

P3 = 99.4MW, PArea1 = 700MW P6 = 287.7MW, PArea2 = 1100MW

1 = 17.856 $/MWhr 2 = 16.185 $/MWhr

C1 = 13,677.21 $/hr C2 = 18,569.23 $/hr

Total generation costC=C1+ C2=

32,246.4 $/hr

Step 2: Perform economic dispatch with two areas interconnected

Area 1 Area 2

P1 = 184MW, P2 = 166.2MW P

4 = 590MW, P

5 = 402.7MW

P3 = 54.4MW, PArea1 = 404.6MW P

6 = 402.7MW, P

Area2 = 1395.4MW

= 16.99 $/MWhr = 16.99 $/MWhr

C1 = 8,530.93 $/hr C2 = 23,453.89 $/hr

Total generation cost, C =C1+ C2=31,984.82 $/hr

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Step 3: Calculate interchange power price

Interchange power from Area 1 to Area 2:PI=P

Area2 PArea2= 1395.4 1100 = 295.4MW

Overall saving after interconnected:

CS=C C = 32, 246.44 31, 984.82 = 261.62$/hr

Assume 50-50 split of saving, Area 1 pays to Area 2:

CP =C1 C1 0.5CS

= 13, 677.21 8, 530.93 130.81 = 5, 015.47$/hr

Each area would have $130.81 reduction in operating cost per hour.

Alternatively, perform economic dispatch for separate areas with:

Area 1 with 50MW steps of power interchange (increase) since is higher.

Area 2 with 50MW steps of power interchange (decrease) since is lower.

Iterate until is equal..

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5 Unit Commitment

To commit a generating unit is to turn it on; that is, to bring the unit up to

speed, synchronize it to the system, and connect it so it can deliver power

to the network.

The problem with commit enough units and leave them on line is one of

economics. A great deal of money can be saved by turning units off

(decommitting them) when they are not needed.

Since the load varies continuously with time, the optimum combination of units

may alter during any period. To determine which units of a plant should operate

for a given load is the problem of unit commitment (plant ordering).

Every electricity supply undertaking is normally under obligation to provide

power to its consumers with some degree of reliability. Therefore, it is necessary

to coordinate both the economy and reliability (security) of unit commitment.

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5.1 Example 8

Suppose the following 3 units are available:

Unit 1: Min = 150MW Max = 600MW

C1 = 510 + 7.2P1+ 0.00142P21 MBtu/h


C2 = 310 + 7.85P2+ 0.00194P22 MBtu/h


C3 = 78 + 7.97P3+ 0.00482P23 MBtu/h

with fuel costs: Fuel cost1 = 1.1 $/MBtu F1 = Fuel cost1 C1 $/h

Fuel cost2 = 1.0 $/MBtu F2 = Fuel cost2 C2 $/h

Fuel cost3 = 1.2 $/MBtu F3 = Fuel cost3 C3 $/h

What is the cheapest way to supply a load of 550MW ?

32

To solve this problem, simply try all combinations of the 3 units (23 = 8). For each

feasible combination, the units will be dispatched using the techniques outlined in

previous sections on economics dispatch.

Unit Max Min P1 P2 P3 F1 F2 F3 Total

1 2 3 MW MW MW MW MW $ $ $ cost

0 0

200 50

400 100

600 150 0 400 150 0 3760 1658 5418

600 150 550 0 0 5389 0 0 5389

800 200 500 0 50 4911 0 586 5497

1000 250 295 255 0 3030 2440 0 5471

1200 300 267 233 50 2787 2244 586 5617

It is the cheapest to run the lowest cost unit (Unit 1) by itself.

This leads to the simple idea of merit loading.

This brute-force approach requires large amount of calculation.

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6 Constraints in Unit Commitment

Many constraints can be placed on the unit commitment problem. Each individualpower system, power pool, reliability council and so forth may impose different rules

on the scheduling of units, depending on the generation makeup, load-curve

characteristics and such.

6.1 Spinning Reserve

Spinning reserve (SR) is the term used to describe system reserved generation

capacity, i.e. it is the amount of generation available from all units synchronized

(i.e. spinning) on the system minus the present load and losses being supplied.

PSR =PGen PLoad PLoss

Spinning reserve must be adequately carried so the loss of one or more units

does not cause too far a drop in system frequency.

Reserve allocation to various units has to obey certain rules, usually set by the

regional reliability councils.

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Typical rules: SR largest loaded unit

SR given percentage of forecasted peak demand

and allocated among fast-responding units and slow-responding units

Examples:

1. A generating plant has a policy to keep a 15% reserve for its peak demand.

If a forecast peak demand is 2200MW, the required spinning reserve would

be 2200MW x 0.15 = 330MW.

2. A power plant has 8 unit of 600MW generators. On a winter day, the

maximum demand is 1500MW. In order to cater the demand, 3 x 600MW

units are on. Therefore, a spinning reserve of 600MW is needed because the

largest or heavily loaded unit is 600MW.

Reserves should also be spread around the power system so as not to bottled

into a particular region due to transmission limitations if a large set is lost in the

another area bottling of reserves. Under severe condition, reserves can run

as islands supplying its own area when system failed.

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SR1 SR2(Transmission limit, PTmax)

Region 1 Region 2Interchange, PT

SR1 + SR2 may be globally adequate but if largest set in one region is lost, the

spare transmission available (i.e. 550MW - interchange) cannot meet the

reserve tranfser from the other region.

e.g.SR1 = 860MW, SR2 = 450MW, PT= 160MW, PTmax = 550MW

Though the entire system has ample reserve (SR1+SR2=1310MW), lost of a

1000MW generator in Region 2 would require a total power transfer of 710MW

from Region 1 and produces a bottling of reserve.

Beyond spinning reserve, the unit commitment problem may involve various

classes of scheduled reserves or off-line reserves such as diesel or

gas-turbine units, hydro-units and pumped-storage hydro-units.

It should also be noted that rules for spining reserve differs from one power

company to another power company as well as in different countries.

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6.2 Thermal Unit Constraints

Thermal generator usually requires an up time of several hours to bring the boiler

temperature and turbine speed up to the full operational conditions before it can be

synchronized and bring on-line.

1. Minimum up time once the unit is running, it should not be turned off

immediately.

2. Minimum down time once the unit is decommitted, there is a minimum time

before it can be recommitted.

3. Crew constraints if a plant consists of two or more units, they cannot both be

turned on at the same time since there are not enough crew members at attend

both units while starting up.

4. Start-up cost it is the cost required to bring a thermal unit on-line. The start-up

cost can vary from a maxium cold-start value to a much smaller value if the unit

is already close to operating temperature.

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37

5. Cooling and banking there are 2 approaches to treating a thermal unit during

its down period. The first is referred as cooling. It allows the units boiler to cool

down and then heat back up when recommitted. The second called banking

requires that sufficient energy be input to the boiler to just maintain operating

temperature. The costs for the two can be formulated as:

Start-up cost when cooling = Cc(1 e

t ) + Cf

Start-up cost when banking = Ct t+ Cf

Banking

Cooling

Start-up

cost

Time-dependent start-up costs

tc

where t = time of cooling

= thermal time constant

Cf = fixed cost

Cc = cold-start cost

Ct = cost of maintaining unit

at operating temperature

Note: up to certain hours, the cost of banking will be less than the cost of

cooling, i.e. hot-start cost < cold-start cost

38

6.3 Must Run

Some units are given a must-run status for stability or security reasons or for such

purpose as supply of steam for uses (e.g. heating) outside the plant itself.

6.4 Fuel Constraints

Some units have limited fuel (for example due to late delivery, bad weather, high

price, forecast error, etc.) or else have constraints that require them to burn a

specified amount of fuel in a given time (for example due to contractual reasons).

7 Unit Commitment Solution Methods

The commitment problem can be outlined as:

There are Nunits to commit and dispatch and Mload levels.

The operating limits on the Nunits are that any unit can supply the individual

loads and that any combination of units can supply also supply the loads.

The total number of combinations is therefore(2N 1)M (all generators off is

excluded) and will grow to a very large number as N and M increased.

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39

7.1 Priority List

Optimum Combination

Load Unit 1 Unit 2 Unit 3

1200

1150

1100

1050

1000

950

800

750

700

650

600

550

500

For a optimised system, units must be shut

down as the load goes down and then

recommitted as it goes back up. A shut-

down rule can be obtained for Example 8

simply by a brute-force technique wherein all

combinations of units will be tried for each load

value taken in steps of 50 MW.

4 PM 4 AM 4PM

600MW

1200MW Unit 3

Unit 2

Unit 1

Unit 2

Unit 3

Time of day

Total load

load-curve

40

The shut-down rule is quite simple: When load is above 1000 MW, run all 3 units;

between 1000 MW and 600 MW, run units 1 and 2; below 600 MW, run only unit 1 .

Alternatively, this rule can be expressed as a priority list of units: 1, 2, 3. It shows

that unit 1 has the top priority to be committed while unit 3 has the lowest priority.

As the above priority list is derived by brute-force, it is very time consumming for a

large system. Simpler methods for the production of approximate priority lists have

been used by utilities. For the above example a simpler scheme is outlined below:

First derive the full load average production cost (FLAPC) table for all units as

follows and then a commitment scheme based on FLAPC can be drawn.

Unit FLAPC ($/MWh) Combination Min MW Max MW

1 9.79 2+1+3 300 1200

2 9.40 2+1 250 1000

3 11.19 2 150 400

The priority order based on FLAPC is 2, 1, 3.

Now the above merit order loading scheme is used.

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41

7.2 Shut-Down Algorithm

The following simple shut-down algorithm can be used together with the priority list

when load is falling.

a) Will dropping the next unit leave enough generation on-line to meet demand

plus reserve requirements? If NO, operate as at present, if YES, goto (b).

b) Determine when the unit will be needed next (when load rises),

say H hours later.

c) If this is less than the minimum shut down time, keep the unit on-line,

if not goto (d).

d) Calculate two costs. First is cost for next H hours with unit on-line; second is

without unit (to choose between banking and cooling depending on cost).

If there is a saving, shut down the unit; otherwise, operate as at present.

e) Repeat this entire procedure for the each unit on the priority list.

42

7.3 Example 9

A supply system consists of four generators, each of which has a rating of 500MW

and a minimum permitted output of 100MW. The operating costs of the units are

given by the quadratic formula:

C=a + bP+ cP2

where Cis in $/h and P in MW. The coefficient values are tabulated below:

Unit a b c

1 500 7 0.02

2 1000 9 0.03

3 500 6 0.03

4 500 4 0.05

i) Determine the full load average production cost of each generator.

ii) Prepare a Priority List for generator loading and indicate the load MW range

over which each row of the list is valid.

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43

iii) Given that

a) load curve :

1 2 3 4 5 6 7 8 hr

400

800

1200

1600

MW

b) spinning reserve is set to 10% of the total generation MW capacity on-line.

c) starting cost when banking :Ctt k$ where Ct = 0.8

starting cost when cooling :Cc(1 et/d) k$ where Cc = 5 & d = 3

d) minimum down time : Unit 1 2 3 4

Hours 4 4 2 2

Determine a full unit commitment schedule for this system.

44

Solution:

i) G1: FLAPC= C1(500)

500 =

500 + 7(500) + 0.02(500)2

500 =18 $/MWh

G2: FLAPC= C2(500)

500 =

1000 + 9(500) + 0.03(500)2

500 =26 $/MWh

G3: FLAPC= C3(500)

500 =

500 + 6(500) + 0.03(500)2

500 =22 $/MWh

G4: FLAPC= C4(500)500

= 500 + 4(500) + 0.05(500)2

500 =30 $/MWh

ii) Priority list: Merit order loading scheme:

Unit FLAPC ($/MWh) Priority Combination Min MW Max MW

1 18 1 1+3+2+4 400 2000

2 26 3 1+3+2 300 1500

3 22 2 1+3 200 1000

4 30 4 1 100 500

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45

iii) Full unit commitment schedule:

Hour Load Load+SR Unit Schedule Remark

1 1600 1760 1+3+2+4

2 1400 1540 1+3+2+4 G4: On to meet SR requirement

3 1200 1320 1+3+2 G4: Off for 4 hrs in banking mode

4 800 880 1+3+2 G2: On (minimum 4hr down time)

5 800 880 1+3+2

6 1200 1320 1+3+2

7 1400 1540 1+3+2+4

8 1400 1540 1+3+2+4

Should G4 be banked or cooled?

Banking cost = (0.8)(4) = 3.2 k$ cheaper option

Cooling cost = 5(1 e4/3) = 3.68 k$

G4 should be banked.

46

8 Other Types of Units

The systems treated so far have included only fossil-fuel generating units. In

practice, however, systems have a mix of different types of units including fossil-fuel,

nuclear, pumped-storage hydro, hydro and other types such as wind generators,

diesel and gas-turbine units, etc.

8.1 Nuclear Units

Although the fixed costs of a nuclear unit may be high, their operating cost are

low due to inexpensive nuclear fuel.

As such, nuclear units are normally base-loaded at their rated outputs. That is,

the reference power settings of turbine-governors for nuclear units are held

constant at rated output.

Nuclear units, therefore, become large in size and they are committed to supply

only base loads and do not normally participate in economic diapatch.

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47

8.2 Pumped-Storage Hydro Units

Pumped-storage hydro is a form of energy storage and is the only practicalmethod of storing electric energy on a large scale.

During off-peak hours these units are operated as sychronous motors to pump

water to a higher elevation. Then during peak-load hours the water is released

and the units are operated as synchronous generators to supply power.

Economic operation of the power system is improved by pumping during

off-peak hours when the system (generation cost) is low, and by generating

during peak-load hours when is high. However, coordinations are needed forincorporating pumped-storage into economic dispatch of fossil-fuel units.

Apart from demand regulation, a large proportion of the spinning reserve is

commonly put on the pumped-storage hydro units because of its fast pick-up.

As a results, pumped-storage plant are often not included in economic dispatch

analysis.

48

8.3 Hydro Units

The input-output relation of a hydro unit is a function of both discharge rate and

effective head.

Hydro units do not have any fuel cost.

They can follow rapid load changes more easily than steam units.

Further, hydro units can be committed within minutes while the commitment of

steam units requires several hours.

The last two features make hydro plants specially suitable for cold reserve and

for working as peaking and regulating plants.

However, while the fuel-supply is not a limiting factor in the operation of thermal

plants, the operation of hydro-electric plants is dependent on the availability of

water flow.

In other words, energy output (MWh) is fixed and the MW dispatch is regulated

by adjusting the operating hours in the case of hydro-elecric plants.

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THE HONG KONG POLYTECHNIC UNIVERSITY

Department of Electrical Engineering

EE4031 Power Systems

Tutorial on Economic Dispatch and Unit Commitment

1. The fuel-cost curves for three generating units are given as follows :

2

1 1160 12 0.025C P= + + 1P $/h 110 120P MW2

2 2160 17 0.050C P= + + 2P $/h 210 100P MW2

3 3180 15 0.010C P= + + 3P $/h 310 80P MW

whereP1,P2andP3are in MW. Neglecting transmission losses, the three generators are

being dispatched economically.a) Obtain by calculation the dispatch of a load of 220MW.

b) Plot the graph of incremental cost versus load from 30MW to 300MW and hencedetermine the incremental cost for a load of 250MW.

2. A two-bus system which has a total load of 400MW and no generator limits is shown inFig.1. The incremental fuel cost functions of the generators and the loss formula of thetransmission link are given below:

1 10.015 50IC P= + $/MWh

2 20.015 50IC P= + $/MWh2

20.0008( 100)LP P= MW

whereIC1andIC2are the incremental costs of generator 1 and 2 when their outputs areP1andP2, respectively.PLis the power loss in the transmission link.

a) Neglecting transmission losses, find the optimal generation for each plant and the

system .

b) Using the solution of (a) as a starting point, find the optimal generation for eachplant and the transmission loss.

c) If the transmission loss is included but not coordinated,i. how much will the total generation cost increase?

ii. what will be the transmission losses?

Fig.1

EE4031, KWCn, revised on 13 September 2010 1


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3. A small power system with total load of 800MW is served by three generating unitswhich have the following cost characteristics.

2

1 15.0 7.0 0.004C P= + + 1P

2P

3P

$/h2

2 26.0 8.0 0.006C P= + + $/h whereP1,P2andP3are in MW

23 37.0 9.5 0.008C P= + + $/h

a) Determine the optimal economic dispatch of generation and the marginal cost ofsupplying an additional kWh of electricity.

b) If the transmission loss associated with generation is given by

2

30.005LP P= MW

Using the results obtained in (i) as the initial estimates, determine the optimaleconomic dispatch and hence obtain the corresponding transmission loss and theaverage system electricity production cost in $/kWh.

4. A supply system consists of four generators. The operating costs of the units are given bythe following quadratic formula :

2C a bP cP = + +

where a, b and c _are constants whilst C and P are in $/hr and MW, respectively. Thecoefficient values and loading limits are tabulated below.

UnitLoading Limits Operating Cost Parameters

Minimum(MW)

Maximum(MW)

a($/hr)

b($/MWhr)

c($/MW_hr)

1 125 550 500 7 0.02

2 100 450 900 9 0.03

3 75 500 400 5 0.03

4 75 400 500 6 0.05___

a) Determine the full load average production cost (FLAPC) of each generator.

b) Determine the priority order for the generators based on the FLAPC, and henceprepare a merit order loading scheme with information on both the unit combinationand load MW range.

c) Comment on the optimality of this scheme and suggest a better approach to theproblem.

5. A power system consists of four generators with the characteristics tabulated below.

Unit MeritOrder

Loading Limits Start-up Cost Coefficients

Minimum(MW)

Maximum(MW)

Cb($/hr)

Cc($)

(_hr)

1 100 500 1100 5500 5

2 150 600 1300 5000 4

3 100 500 1200 5500 34 150 600 1300 6000 4



27/31

The units are scheduled according to a strict merit order with the following system andoperation constraints:

a) The spinning reserve requirement is 10% of the system loading MW capacity.

b) The restarting cost after banking and cooling for all generators are given by

and

bC t

/(1 )tcC e , respectively, in dollars, where tis the time of cooling in hours.

c) The minimum down time for a generator taken off-line is 2 hours for all units.

The system demand over a ten hours period is given in the table below with theassumption that the load demand remains constant for each hour.

Hour 1 2 3 4 5 6 7 8 9 10

Load (MW) 1100 800 1050 1200 1500 900 900 1200 1700 1800

Determine a full unit commitment schedule for the above system including specification

of which generator, if any, should be banked or cooled, when taken off.

6. A supply system consists of four generators, each of which has a rating of 500MWand aminimum permitted output of 50MW. The operating costs Coand the restarting cost Crafter banking and cooling of the units are given by the following formulas.

2

oC a bP cP = + +

/

banking

(1 ) cooling

b

r t

C

C tC

C e

=

where Co, Cr,Pand tare in $/hr, $,MW, and hr, respectively, and the coefficient valuesare tabulated below.

Unit a b c Cb Cc 1 500 7 0.02 1100 5500 5

2 1000 9 0.03 1300 6000 3

3 600 6 0.03 1200 5500 2

4 700 4 0.05 1300 5000 4

a) Determine the full load average production cost of each generator.

b) Prepare a priority list for generator loading and indicate the load MW range overwhich each row of the list is valid.

c) If the spinning reserve requirement is 10% of the system loading MW capacity andthe minimum down time is 2 hours for all units, determine the full generatorschedule for the following system demand over a nine hours period.

Hour 1 2 3 4 5 6 7 8 9

Load (MW) 1300 800 950 1800 1700 850 800 950 1300



28/31

EE4031 Power Systems

Tutorial Solution on Economic Dispatch and Unit Commitment

1.

a) IC1= 12 + 0.05P1 10


29/31

2. a) As generator 1 and 2 has the same incremental cost function with no transmissionlosses, generator 1 and 2 share the load equally, i.e.

P1= 200 MWP2= 200 MW

= 0.015(200)+50 = 53 $/MWh

b) L1= 1.0, L2=2

2

1 11.16 0.0016

1 LP P

P

=

1 1 10.015 50L IC P = + = 150

0.015P

=

22 2

2

0.015 50

1.16 0.0016

PL IC

P

+= =

2

1.16 500.015 0.0016

P

=+

1 2 (300 100) LP P P+ = + +

( )2

50 1.16 50 1.16 50400 0.0008 1000.015 0.015 0.0016 0.015 0.0016

+ = + + +

( )2

1.16 50 1.16 50400 0.0008 100 0.015 500.015 0.0016 0.015 0.0016

= + + + +

Substitute = 53 $/MWh and iterate: 0= 53.00 $/MWh hence, P1= 274.73 MW1= 54.28 $/MWh P2= 125.80 MW2= 54.10 $/MWh PL= 0.53 MW3= 54.12 $/MWh

c) i. Loss included but not coordinated: 1 2 1 2IC IC P P = =

Hence, (1) becomes: MW22 2 1 22 400 0.0008( 100) 204.36P P P P= + = =

Therefore, cost increase274.73 125.8

1 1 2 2

204.36 204.36

IC dP IC dP= + = $371.2/h

ii. Transmission loss, PL= 0.0008(204.36-100)2= 8.713 MW

3. a) IC1= 7.0 + 0.008P1$/MWh PD= P1+ P2+ P3 = 800MWIC2= 8.0 + 0.012P2$/MWhIC3= 9.5 + 0.016P3$/MWh

The required optimal condition is when= IC1= IC2= IC3

Hence P1= 479.81 MW and = 10.84 $/MWh = 1.084 $/kWh

P2= 236.54 MWP3= 83.65 MW

b) L1= 1.0, L2= 1.0, L3=3

3

1 11 0.01

1 LP P

P

=

1 1 17.0 0.008L IC P = + = 17.0

0.008P

=

2 2 28.0 0.012L IC P = + = 2 8.00.012P =



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32 2

3

9.5 0.016

1 0.01

PL IC

P

+= =

3

9.50.016 0.01

P

=+

1 2 3800

LP P P P+ + = + where PL= 0.005P3

2

( )2

7.0 8.0 9.5 9.5800 0.0050.008 0.012 0.016 0.01 0.016 0.01

+ + = ++ +

( )2

9.5 9.5800 0.005 0.000096 0.148 500.016 0.01 0.016 0.01

= + + + +

Use = 108.38 $/MWh as the initial estimate and iterate: 0= 10.84 $/MWhhence, P1= 522.63 MW 1= 11.19

P2= 265.08 MW 2= 11.18P3= 13.15 MW 3= 11.18PL= 0.86 MW

Total production cost = C1+ C2+ C3= 7,437.53 $/h

average production cost = 0.93 c/kWh

4. a)2

1

500 7 550 0.02 55018.91

550C

+ + = = $/MWh

2

2

900 9 450 0.03 45024.5

450C

+ + = = $/MWh

2

3

400 5 500 0.03 50020.8

500C

+ + = = $/MWh

2

4500 6 400 0.05 400 27.25

400C + + = = $/MWh

b) Generator Priority orderG1 1G2 3G3 2G4 4

Loading limitsLoading units Min Max

G1+G3+G2+G4 375 1900G1+G3+G2 300 1500G1+G3 200 1050G1 125 550

c) Scheme based on the priority ordering is not optimum as only average costs areconsidered. A better approach would be one based on dynamic programming (DP)using detailed production cost and economic dispatch analysis.



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5. Hour Load Load+SR Unit Schedule1 1100 1210 1+2+32 800 880 1+2+3 Unit-3: On (min 2hr down time)3 1050 1155 1+2+34 1200 1320 1+2+3

5 1500 1650 1+2+3+46 900 990 1+2 Unit-3: Off for 2 hrs Banking7 900 990 1+28 1200 1320 1+2+3 Unit-4: Off for 3 hrs Cooling9 1700 1870 1+2+3+410 1800 1980 1+2+3+4

Hr: 6-7, Unit-3: Banking = $12002 = $2400 CheaperCooling = $5500(1 e-2/3) = $2676.2

Hr: 6-8, Unit-4: Banking = $13003 = $3900

Cooling = $6000(1 e-3/4) = $3165.8 Cheaper

6. a) Unit FLAPC($/MWh)1 182 263 22.24 30.4

b) The FLAPC based priority order for the these units is as follows:

Unit $/MWh Min MW Max MW1 18 50 5003 22.2 100 10002 26 150 15004 30.4 200 2000

c) Hour Load Load+SR Unit Schedule1 1300 1430 1+3+22 800 880 1+3+2 Unit 2: On (min 2 hrs down time)3 950 1045 1+3+24 1800 1980 1+3+2+4

5 1700 1870 1+3+2+46 850 935 1+3 Unit 2: Off for 2 hrs Banking7 800 880 1+3 Unit 4: Off for 4 hrs Cooling8 950 1045 1+3+29 1300 1430 1+3+2

Hr 67, Unit 2: Banking = $1,3002 = $2,600 cheaper choice

Cooling = $6,000(1e-2/3) = $2,919.5

Hr 69, Unit 4: Banking = $1,3004 = $5,200

Cooling = $5,000(1e-4/4

) = $3,160.6 cheaper choice

Documents

EE4031 2 Economic Operation