Lecture 12 Economic Optimisation of Power System O

7/31/2019 Lecture 12 Economic Optimisation of Power System O

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Economic Optimisation of Power system Operation

The extensive interconnection of power systems has made the operation of a power

system, in the most economic way, a complex subject. It is intended here only to

outline the problem and indicate the types of approaches needed.

Optimisation can be considered in a number of different ways according to the

timescale involved (hourly, daily, yearly etc.). The main considerations are:

Accurate load forecasting

Economic scheduling of generators

Economic loading of online generators

Accurate load forecasting

As already discussed, some form of control usually assists the forecasting of system

loads either directly or through tariff schemes.

Economic scheduling of generator loads

To determine the economic distribution of load between the various generating units,

the variable operating costs of the unit must be expressed in terms or its power output.

Let Fi(Pi) be the total operating cost per hour (including fuel) of a unit as a function of

output power Pi of the unit i. It can be shown that the most economic operation of a

system (ignoring transmission loss) is when all the incremental costs of each power

unit are identical. The incremental costi of a unit i is given by:

i

iii

P

PF

)( (1)

This can be explained by considering a two unit system. A change in overall cost for

running the system Ftotal will occur if the output powers of the two units change and

this is given by

221122

221

1

11 )()( PPPP

PFP

P

PFFtotal

(2)

where Pi is the change in unit i output power. If the total output power remainsunchanged then PPP 21 giving

PFtotal )( 21 (3)

Thus if the incremental costs are not identical then a reduction in overall costs can

always be achieved (Ftotal negative) by adjusting the units loading (Pnon-zero but

positive or negative). An important consequence of this is that each unit will not

necessarily be operating at its most efficient output power. The following are typical

generator unit parameters.


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Typical generator F costs in per pu of two machines as a function of their output

power are:

/hr96.05.1)(2

111 PPF (4)

/hr64.0)( 22

222 PPPF (5)

The cost curves are depicted in Figure 1.

Typical cost curves

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

P /pu

F

//hr

Figure 1 Running cost of each unit as a function of their power output

The Efficiency of the unit is then given by Output Power divided by cost For

Phr/96.05.1

1Efficiency

11

1

PP

(6)

Phr/64.01

1Efficiency

22

2

PP

(7)

These are depicted in Figure 2 have a maxima at 0.8 pu which is quite typical.

1

2


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Efficiency curve

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

P /pu

Efficiency

/Phr/

Figure 2. Efficiency of each unit as a function of their outputs

On the other hand the incremental costP

F

is given by

/Phr3 11 P (8)

/Phr12 22 P (9)

If the total output power Ptotal is 2 Pu then as P2=Ptotal-P1=2-P1 then equation (9)

becomes

/Phr124 12 P (10)

The incremental cost curves as a function of Unit 1 output are then as given in Figure

3.

1

2


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Increm ental cost curve againt unit 1 powe r output

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

P1 /pu

F

//Phr

Figure 3 Incremental cost curves as a function of Unit 1 output

In the two unit system the most efficient operating point is where the two incremental

cost curves intersect. Which is when both machines are output power of 1 pu . If unit

1 has an output less than 1 pu then increasing its load at the expense of Unit 2 will

decrease the overall running costs as 12. The total system cost curve F1+F2 as a function of

Unit 1 output power is given in Figure 4. showing the minimum at 1 pu

Total cost curve

0

1

2

3

4

5

6

7

8

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

P1 pu

F

//hr

Figure 4. Total cost of 2 unit system as a function of unit 1 output

1

2


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Example

The incremental costs in pounds per Megawatthour for a two unit plant is:-

11 1

1

0.004 4.0GG

fP

P

22 2

2

0.006 3.2GG

fP

P

The total load varies from 200-1200 MW. Each machine can only operate at an output

between 100-600 MW. Find the allocation of load between the units for minimum

costs for the full range of loads.

Solution:

First consider the minimum loading condition (200 MW total or 100 MW each). Theincremental costs are:

1 4.4 per MW hour

2 3.8 per MW hour

Thus initially, for low loads increasing from minimum load, Unit 2 can be

preferentially loaded up until its units cost equals that of unit 1. The minimum load

where the two units incremental costs are equal will be when

1 2 4.4 per MW hour which is when:

1 2 1 2100 MW, 200 MW and 300 MWG G total G GP P P P P

At the maximum loading (1200 MW or 600 MW each) the incremental costs are:

1 6.4 per MW hour

2 6.8 per MW hour

Thus initially as the total load is reduced from the maximum Unit 1 should be

preferentially loaded (kept at 600 MW) until both incremental costs are the same (i.e.

1 2 6.4 per MW hour ) and this occurs when

1 2 1 2600 MW, 533 MW and 1133 MWG G total G GP P P P P

For total loads between 300 MW-1133 MW the two units can operate at the optimum

condition 1 2 Thus we can derive the loading for each of the units in the following

way:

1 2 1 2, total G GP P P


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Thus

1 2

1 1

1

2

0.004 4 0.006 3.2

0.004 4 0.006( ) 3.2

0.6 80 MW

and 0.4 80 MW

G G

G total G

G total

G total

P P

P P P

P P

P P

The resulting graph is:

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Lecture 12 Economic Optimisation of Power System O