Power system reinforcement through reliability and revenue considerations

Short paper Power system reinforcement through reliability and revenue considerations A h m e d M H Rashed Department of Electrical Engineering. University. Alexandria. Egypt

Alexandna

Madiha M Awad Alexandna DIstnbutJon Co, Alexandna. Egypt

The problem of power system reinforcement is addressed from the point of view of overall system reliability and revenue only. A probabilistic function representing the loss of revenue due to service interruption is proposed as a measure of the global performance of the system and is taken as a criterion for reinforcement. A technique based on the sensitwity of the proposed function to the forced outage rates of system components is proposed to carry out the reinforcement. Examples are given to illustrate the technique. They show that parallel lines are not, in general, the best reinforcement. Keywords: reliability, composite planning reliabdity calculations

system

reliability,

I. I n t r o d u c t i o n The reliability of power systems decreases with time, partly due to aging of components resulting in increased outage rates, and partly due to increased demand causing overloading of components which in turn decreases their reliability1. Other factors, such as poor design or poor reliability of protectzon systems, may also lead to decreased power system reliability2. Accordingly, power systems often require reinforcement. In transmission and distribution networks, reinforcement is usually carried out ezther by renewing old lines or by adding new lines. The general practice is to add a new line in parallel with the old one. In this paper the reinforcement of transmission and distribution networks is addressed. An innovative technique for reinforcement based solely on considerations of reliability and loss of revenue is presented. It will be shown that the addition of parallel lines is not always the best course of action. Although the choice of reinforcement is also governed by other factors, such as the cost of reinforcement and operational considerations (e.g. short-circuit levels, etc.), these are not considered in this paper. Received 31 May 1985, Revised 15 April 1987

Vol 10 No 1 January 1988

II. Probabilistic revenue loss f u n c t i o n A direct result of decreased reliability of transmission and/or distribution networks Is an increase in the duration of service interruption at load points. This is represented by the average annual customer interruption rate (AAIR) 3. It is a good measure of the continuity of supply at any load point, but not as good for measuring the performance of the overall power network. In other words, individual AAIRs indicate whether the power system is performing adequately or inadequately as far as individual load points are concerned. They give little or no indication on how adequately the overall power system is performing. Therefore AAIRs alone cannot be taken as a base for network reinforcement except in the case when reinforcement is required to serve a particular load point. It is proposed that a good judgement of the adequacy of the overall performance of a power network can be obtained by constructing a 'probabilistic revenue loss function'. This is a function of the weighted loss of revenue to the power company resulting from discontinuity of supply at all load points. It is suggested in the form

F= ~ (AAIR)IDlK31(K,I+K21) 1=1

unit money/year

(1)

where (AAIR)I average annual interruption rate at load point i in hours/year q average yearly power demand at load point i Kzl price of unit energy at load point i K21 penalty factor in unit money/unit energy of interruption at load point i, which represents the penalty payable by the power company for service interruption K31 pnority factor assigned to the load at load point to reflect its importance total number of load points

0142-061 5/88/010059-04/S03.00© 1988 Butterworth Et Co (Pubhshers) Ltd

59

Since the AAIRs are dependent on the forced outage rates of network components, the probabilistlc revenue loss functmn can be rewritten in the form F= f(Pt: l= 1,2,...,m)

(2)

as follows: 1

Evaluate Fr and F, to check whether or not the system requires reinforcement as given in Section III

2

If the system requires reinforcement (Fa > Fr), evaluate the partial derivatives of F with respect to the FORs of system components at their actual values. This gives the sensitivities 'a~' of F to the FORs, i.e.

where P~ is the forced outage rate of component I and m is the total number of components of the network.

at = O F / O P t

III. Reinforcement criterion Standard values for the forced outage rates (FORs) of the different components of the power system may be considered to form a basis for the judgement of the performance of individual components from a rehabtlity point of view. Other values representing the minimum acceptable reliability requirements may also be used as seen fit by the power company in considering its operating standards and the prevailing conditions. It is proposed that the standard (or recommended) values of the FORs of system components can well be used for a reliabihty judgement of the performance of the overall system. This is achieved by evaluating the probabilistic revenue loss function F of equation (2) for these values, resulting in a base or reference value Fr of F: Fr= f(Pi, : l = 1,2 . . . . m)

(3)

where Pls t is the standard value of the F O R of component I. The actual performance of the system is measured by evaluating F for the actual values of the FORs of system components obtained from available statistical data. The value of the function in this case is denoted by Fa, I.e. F R = f ( P ~ , c : l = 1,2 . . . . . m)

(4)

where P~c, is the actual value of the FOR of component 1. Comparison of Fr and Fa of (3) and (4) determines whether or not the system needs reinforcement. If F, is greater than Fr, reinforcement is needed. This is justified on the basis that it means that there will be a possible loss of revenue in excess of the maximum loss acceptable to the power company If on the other hand F, 1s less than Fr, reinforcement is not, in general, needed. In this case reinforcement may still be needed to serve a particular load point as can be determined via the individual FORs of system components. However, since the focus here is upon the global performance of the overall system, the condition that F, is less than Fr is considered to mean that no reinforcement is needed.

l = 1,2 ....

(5)

,m

3

The component for which a t IS largest is the component whose P~ has the most effect on F. It is the logical candidate for reinforcement.

4

Suggest a reinforcement and re-evaluate F, with the reinforcement included.

5

If F, is still larger than Fr, repeat steps 2-4 with previous reinforcements included.

6

When F, becomes less than F,, the process is terminated.

Steps 4 - 6 can be repeated for other reinforcements of the component picked in step 3 the first time. This results in several schemes of reinforcement which can be compared as to their cost, their effect on system operation, etc. to arrive at a final decision.

V. Examples Two examples will be given to illustrate the technique presented in this paper. The first example is a simple radial system, while the second is a five-bus ring system.

V 1 Example 1 Consider the ra&al system shown in Figure 1. Power is fed at bus A and is delivered to loads at buses B and C through three lines. Recommended and actual FORs of the three lines are given in Table 1. The load parameters are given m Table 2. The probabilistic revenue loss function is given by F = 106(23.792P1 P2 + 1.927P3)

A

B

(6)

5 LoadB ~Load C

4[a)

C

4(b)

IV. Reinforcement technique If the power system does require reinforcement, it must be carried out to bring about a reduction in Fa to within the limiting value Ft. To cut down the number of reinforcements, this reduction must be obtained in the most effective way. Therefore the logical choice for the reinforcement is reinforcement of the system component which has the most effect on Fa. This can be achieved through investigation of the sensitwlty of F a to the FOR of each component. A scheme can therefore be proposed

60

Figure 1. System of Example 1 Table 1. Recommended and actual FORs of lines of system of Example 1 Line

Recommended FOR

Actual FOR

1, 2 3

1 2 x 10 - 3 0.25 x 10- 3

1 37x 10 - 3 0.274 x 10- 3

Electrical Power 8- Energy Systems

Table 2. Load parameters of system of Example 1 D (MW)

Load B C

Kt (£/kWh)

16 11

0.03 0.02

function F will be given by F = 106(23.792PIP2P4 + 21.865PtP2P 3

K 2

(£/kWh) 0.1 0.0

K3

(8)

+ 1.927/3/'4) giving

1.2 1.0

Fa=0.953

£/year

i.e. Fa becomes extremely small.

Q

It is evident that alternative (b) is more effectave than the usual alternaUve (a). However, other factors, such as length, capacity, etc. and their effect on cost, should be considered before deciding which alternative is to be used.

() E 9(c)

9(0)

" ' - .,, 9 (b)

2

V.2 E x a m p l e 2 The system of Figure 2 is considered. Each of the two generators is assumed to be incapable of supplying all loads by itself. Therefore line 3 is necessary in case of an outage of lines 1 and 6 or lines 2 and 7. Accordingly line 3 (or any additional parallel line) may be considered as a reinforcement of either set. By the same token, a line connected between buses B and D is a reinforcement of lines 1 and 6, and a line connected between buses A and E is a reinforcement of hnes 2 and 7. Tables 3 and 4 give the FORs of the system lines and the load parameters respectively. The function F is given by

lO(aand b)

B Lood B

Load C

Figure 2 System of Example 2

F = 107{68.801P~P6[P3 +P2P7 + P4(P5 +Pa)]

Hence

+ 4.906P2P7 [P3 + P, P6 + P4(P5 +/'8)] F,= 516.061

£/year

+ 15.33[P2P7 + Ps(P4 +/:'5)]

and

x [P,P6+P5(P4+Ps)]} (9) Hence

F,=572.782

£/year F,=1971.048

Clearly F , > F , . The system therefore requires reinforcement. Sensitivities o f F to Pt, P2 and P3 for actual values of FORs are given by

£/year

and Fa = 10 398.926

£/year

al = ~F/OPt = 32.595 x 103 a 2 = c3F/c~P2 = 32.595 x 103

Table 3. Recommended and actual FORs of lines of system of Example 2

a 3 = ~ F / ~ P 3= 1.927 x 106

Therefore line 3 is the line to be reinforced. This is carried out by adding a fourth line whose FOR is assumed to be 0.0014. Two alternatives are possible: (a)

Line 4 is connected in parallel with hne 3. In this case F = 106(23.792PtP2 + 1.927P3Pa)

(7)

Actual values of P1, P2 and P3 and the assumed value of/'4 give F, = 45.524

£/year

i.e. Fa becomes much smaller than reinforcement is thus sufficient. (b)

Recommended FOR

Actual FOR

I, 6 2, 7 3 4, 5, 8

7.5x 10 -3 1.0x 10 -2 4.5x i0 -2

8.565 x 2 855x 1.142 x 0.571 x

5 . 0 x 10 - 3

10 -3 10 -z 10 -1 10 -2

Table 4. Load parameters of system of Example 2

Load

D

(MW)

K1

K2

( £ / k W h ) (£/kWh)

K3

0.04 0.04 0.05

1.1 1.0 2.0

Ft. This

Line 4 IS connected between buses A and C. In this case line 4 will help feed both loads C and B. The

Vol 10 No 1 January 1988

Line

A B C

297.5 140.0 35.0

0.2 0.0 0.2

61

The sensitivities of F in (9) are calculated as

parallel with lines 2 and 7 The function F will have the same form as equation (10) except that the product P2P7 is replaced by the product P2PTPlo The value of Fa becomes

a 1 = OF/t3P1 = 679.645 x 103 a 2 = 8 F / ~ P 2 = 162.184 × 103

Fa = 49.909

£/year

a a = OF/OP3 = 90.458 x 10 3

which is clearly less than Fr. a,,=OF/SP4=

1.925 x 10 a

a 5= 8F/SP5 =

2.179 x 103

(b) The second alternative is an additional line (line 9) connected between buses B and D. The same procedure reveals that an additional reinforcement is also needed for lines 2 and 7. Line 10 is used in parallel with lines 2 and 7, resulting in a final value of Fa of 160.005 £/year.

a 6 = OF/OP6 = 679.645 x 103 a,7 = 8F/OP7 = 162.184 x 103

(c) a~=OF/t3P 8=

1.530 x 103

Therefore the parallel combination of lines 1 and 6 should be reinforced. Three alternatives are possible. (a)

Addition of a new line (line 9) in parallel with lines 1 and 6. The new line is assumed to have a F O R of 0.004 568. The function F becomes F = lO7{68.801P, P6P9[P3 + PEP7 + P4(P5 + P8)]

In the third alternative line 9 is connected in parallel with line 3, resulting in Fa assuming the value of l15.86£/year, which means that this is the only reinforcement required.

Comparison of the three alternatives must be carried out as to cost particularly, since alternatives (a) and (b) require two lines while alternative (c) requires only one. However, from a reliability point of view, the example shows that reinforcement of lines 1 and 6 is not best obtained by a parallel line as generally practised.

+ 4 906P2Pv[P 3 + P1P6P9 + P4(P5 + Ps)] + 15.33[PEP7 + Pa(P4 + Ps)]

× [P, P6Pg+P~(P4+Ps)]}

VI.

(10)

Thus F~ = 4604.361

£/year

i . e . F , is still larger than F~. An additional reinforcement is therefore required. To determine the location of this additional reinforcement, the sensitivities of F in (10) are calculated as a 1 = 3104.616 a 2-- 160.328 × 103 a 3 =40.216 x 103 a 4 = 1287.216 a s = 1828.155 a 6 = 3104.616

Conclusions

in this paper a probabllistic revenue loss function is proposed to measure the adequacy of performance of the overall power system from the point of view of reliability and service interruption. This probabilistic function is then taken as a criterion for system reinforcement. Reinforcement is required when the value of this function, which is dependent on the FORs of all system components, exceeds a calculated standard value. Reinforcement is carried out using a technique based on the sensitivities of the probabilistic function to the FORs of individual components. The examples show the simplicity of the technique. They further reveal that parallel lines, as generally practised, are not always the best answer to the reinforcement problem. It should be noted that for large systems the construction of the probabfllStiC function will be a difficult task, though not impossible. Furthermore, the number of alternative reinforcement plans will be quite large. Therefore in such cases the technique can be applied to localized subsystems and overall system reinforcement can be done piecewlse.

a7 = 160.328 x 103 a s = 1114.954

VII.

References

1 Smith, C O Introduction to Rehabthty m Design McGraw-Hall

a 9 = 5821 158 Thus the additional reinforcement should be for lines 2 and 7. This is carried out by connecting a second line (line 10) with a FOR of 0.004 568 in

62

Kogakusha (1976) 2 Becker, P W and Jensen, F Design of Systems and Circuits for Maximum Rehabthty or Maximum Production Yield McGrawHdl Kogakusha (1977) 3 Billinton, R Power System Rehabthty Evaluation Gordon and Breach Soence Pubhshers, New York (1970)

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