Measurement of reliability parameters for a power

IJRET: International Journal of Research in Engineering and Technology

eISSN: 2319-1163 | pISSN: 2321-7308

MEASUREMENT OF RELIABILITY PARAMETERS FOR A POWER PLANT BY USING ALGEBRA OF LOGICS Reena Garg1 1

Assistant Professor Mathematics, YMCA University of Science & Technology, Faridabad – 121006

Abstract In this paper, the author has considered a power plant to evaluate its reliability parameters by using algebra of logics. The system configuration has been shown in fig-1. In this power plant, there are two identical generators G1 and G2 in standby redundancy. Only one generator is sufficient to fulfill the power requirements of whole system. The author has used four change over devices CO1 through CO4 to feed power to four machine M 1 , M 2 , M 3 and M 4 .It is assumed that the cables used, to connect all equipments, are hundred percent reliable and we need to operate two machines simultaneously. These machines can be operated by switching devices

M 1 , M 2 , or M 3 and M 4

SD1 through SD4 .

Keywords: power plant, identical generators, standby redundancy, changes over devices.

--------------------------------------------------------------------***-----------------------------------------------------------------1. INTRODUCTION In the beginning, the whole power plant is operable and there is no repair facility to repair a failed unit/component of power plant.

The author has obtained the reliability of the system as a whole and reliability function for the system in two cases, i.e., when failure rates follow weibull or exponential time distribution. An important reliability parameter, mean time to failure (M.T.T.F.), has also been evaluated in both the cases. In the end of paper, the author has been given a numerical computation and its graphical illustration to highlight the important results for their practical importance. The results obtained in this study, can be applied for the power plant of similar configuration.

Fig-1: System Configuration

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2. ASSUMPTIONS The following assumptions have been associated with this model: (1) Initially, all the components are good.

(2) There is no repair facility available.

(3) Failure times of all components of the power plant are arbitrary.

(4) Failures are statistically-independent.

(5) Reliability of each component of power plant is known in advance.

(6) Transition of power from one component to other is all reliable.

(7) Every component of the system remains either in good or in bad state.

(8) Only one generator is sufficient to fullfil the power requirement of whole power plant.

(9) Mostly, we need two machines M1 and M2 or M3 and M4 to operate simultaneously.

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 x1 x  1  x1  x1 F x1 , x 2   x10    x  2  x2 x  2  x 2

x3

x4

x3

x4

x3

x4

x3

x4

x3

x4

x3

x4

x3

x4

x3

x4

x5 x 7  x5 x8  x6 x9   x6 x10  x5 x 7   x 5 x8  x 6 x9   x6 x10 

5. SOLUTION OF THE MODEL Using algebra of logics, we may write equation (1) as:

F x1 , x2   x10   x3

x4   f x1 , x2 , x5  , x10 , …(2)

3. NOTATIONS The following notations have been used throughout this paper: : States of generators G ,G x1 , x2 1 2 respectively; : States of change over devices CO1 x3 , x 4 , x5 , x 6 through CO4; : x7 , x8 , x9 , x10 States of switching devices SD1 through

SD4 ;

xi

:

Negation of state x i ,

xi  / Ri

: : :

=1, in good state; =0 in bad state; Conjunction/Disjunction; Reliability of the component corresponding to state x i ;

Si

:

 1  Ri , i ;

RS

:

RSE t  / RSW t 

:

Reliability of the power plant as a whole; Reliability functions when failures follow Weibull/exponential time distribution.

4.

…(1)

FORMULATION

OF

i ;

MATHEMATICAL

MODEL By making use of Boolean function technique, the conditions of capability of the successful operation of the power plant in terms of logical matrix are expressed as:

Where

 x1 x  1  x1  x1 f x1 , x 2 , x5  , x10    x  2  x2 x  2  x 2

x5 x7  x5 x8  x 6 x9   x6 x10  x5 x 7   x 5 x8  x 6 x9   x6 x10 

…(3)

Substituting

B1  x1

x5

x7 

B2  x1

x5

x8 

…(5)

B3  x1

x6

x9 

…(6)

B4  x1

x6

x10 

…(7)

B5  x2

x5

x7 

…(8)

B6  x2

x5

x8 

…(9)

B7  x2

x6

x9 

…(10)

B8  x2

x6

x10 

…(11)

…(4)

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in equation (3), we have  B1  B   2  B3    B4  f  x1 , x 2 , x5  , x10    B   5  B6  B   7  B8   

 x1

B2 B2 B2 B2 B2 B2 B2

B3 B3 B3 B3 B3 B3

B4 B4

 x1 B1 B2 B3    x1

B4 B4 B4

B5 B5 B5 B5

B6 B6 B6

x5 x5

 x1  B1 B2   x1  x1

B7 B7

x5

x6

x5

x6

x9 x7

x6 x9 x6 x7

B1 B2 B3 B4 B5  x1

    …(13)        B8 

 …(15) x9 

x8 x10 x8

x2

 x1 x2 B1 B2 B3 B4 B5 B6   x1 x2  x1 x2

 …(16) x9 x10  x7 

x5

x5 x7 x5 x6

(17)

x8   x7 x8     x7 x8 x9 x10 

x5 x 6

…(18)

 x1 x2 B1 B2 B3 B4 B5 B6 B7    x1 x2

x5 x6 x5 x 6

x9  x7 x8 x9 

…(19)

   x7 

x5

…(14)

x5

 x1 x5 B1 B2 B3 B4    x1 x5

Now we have

 x1 B1   x1  x1

x8 

x7

Similarly, we obtain

…(12)

Using orthogonalisation algorithm, equation (12) may be written as  B1 B  1  B1  B1 f x1 , x 2 , x5  , x10    B  1  B1 B  1  B1

x5

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And

 x1 x2 x5 x6 x9 x10  B1 B2 B3 B4 B5 B6 B7 B8     x1 x2 x5 x6 x7 x8 x9 x10     x 1    x7 

…(20)

x5

x8 

Therefore, equation (13) becomes, subjected to equations (14) through (20):

 x1 x  1  x1   x1 x  1  x1  x 1 f  x1 , x 2 , x5  , x10     x1   x1  x1   x1  x  1  x1   x1

x5 x5 x5

x7 x 7 x6

x8 x9 x 7

x5 x5

x6

x5

x6

x2

x5

x2

x5

x7 x 7

x2

x5

x2

x6

x9 x 7

x8

x9

x10 x8

x9

x 6

x8 x 7

x8

x5 x5

x6

x 7

x8

x9

x6 x6

x8

x9

x2

x5 x5

x9 x 7

x2

x5

x6

x9 x 7

x10 x8

x9

x2 x2

x6

x10

              x10        x10 

…(21)

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Using equation (21), equation (2) gives

 x1 x  1  x1   x1 x  1  x1  x 1 F  x1 , x 2  , x10     x1   x1  x1   x1  x  1  x1   x1

x3

x4

x5

x3

x4

x3

x4

x5 x5

x3

x4

x3

x7 x7 x6

x8 x9 x7

x6

x4

x5 x5

x3

x4

x5

x6

x2

x3

x4

x5

x2

x3

x4

x5

x2

x3

x4

x5

x2

x3

x4

x2

x3

x4

x5 x5

x2

x3

x4

x6

x2

x3

x4

x5 x5

x2

x3

x4

x5

x6

x6

x8

x9

x9 x10 x7 x8 x9

x10

x7 x 7 x8 x 6 x7

x8

x6

x7

x8

x9

x6

x9 x7

x8

x9

x9 x7

x10 x8 x9

x6

              x10        x10 

…(23)

Since, R.H.S. of equation (23) is disjunction of pair-wise disjoint conjunctions, therefore, the reliability of considered power plant will be:

RS  Pr F x1 , x2   x10   1

 R3 R4 R1 R5 R7  R1 R5 R8 S 7  R1 R6 R9 S 5  R1 R5 R6 R8 R9 S 7  S 5 S 9 R1 R6 R10  S 7 S 9 R1 R5 R6 R8 R10  S1 R2 R5 R7  S1 S 7 R2 R5 R8  S 6 S 7 R1 R2 R5 R8  S1 S 5 R2 R6 R9  S 7 S 9 R1 R2 R5 R6 R8 R10  S1 S 7 S 8 R2 R5 R6 R9  S1 S 5 S 9 R2 R6 R10  S1 S 7 S 8 S 9 R2 R5 R6 R10  , Where, S i  1  Ri , i  1,2,  10. or,

RS  R3 R4 R1 R5 R7  R1 R5 R8  R1 R6 R9  R2 R5 R7  R1 R6 R10  R2 R5 R8

 R2 R6 R9  R2 R6 R10  R1 R5 R6 R8 R9  R1 R5 R6 R9 R10  R1 R5 R6 R8 R10  R1 R5 R6 R7 R8 R9 R10  R1 R2 R5 R6 R7 R8  R1 R2 R5 R6 R8 R10  R1 R2 R5 R6 R7 R8 R9 R10  R1 R2 R5 R6 R7 R9  R1 R2 R5 R6 R8 R9  R2 R5 R6 R7 R8 R9  R1 R2 R6 R9 R10  R1 R2 R5 R6 R7 R10  R1 R2 R5 R6 R8 R10  R2 R5 R6 R7 R8 R10  R2 R5 R6 R7 R9 R10  R2 R5 R6 R8 R9 R10  R1 R2 R5 R6 R7 R8 R9 R10  R1 R5 R7 R8  R1 R5 R6 R9  R1 R5 R6 R7 R8 R9  R1 R2 R5 R7  R1 R5 R6 R10

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 R1 R6 R9 R10  R1 R5 R6 R7 R8 R10  R1 R5 R6 R8 R9 R10  R2 R5 R7 R8  R1 R2 R5 R6 R8  R1 R2 R5 R6 R7 R8 R10  R1 R2 R5 R6 R8 R9 R10  R1 R2 R6 R9  R2 R5 R6 R7 R9  R2 R5 R6 R8 R9  R1 R2 R5 R6 R7 R8 R9  R1 R2 R6 R10  R2 R6 R9 R10  R2 R5 R6 R7 R10  R2 R5 R6 R8 R10  R1 R2 R5 R6 R7 R8 R10  R1 R2 R5 R6 R7 R9 R10  R1 R2 R5 R6 R8 R9 R10  R2 R5 R6 R7 R8 R9 R10 

…(24)

a 9  c  1  5  6  7  8  9  10

6. SOME PARTICULAR CASES 6.1 CASE I: If the Reliability of each Component of Power Plant is R:

a11  c  1  2  5  6  7  8

In this case, equation (24) yields:

RS  8R 5  9R 6  R 7  7 R 8  6R 9  2R10 …(25) 6.2 CASE II: When Failure Rates follow Weibull Distribution Let

i be the failure rates corresponding to component

states xi ,  i  1,2    10 . Then, reliability function of power plant, at an instant„t‟, is given by:

25





24



RSW t    exp .  ai t    exp .  b j t  i 1

where,



a10  c  2  5  8



(26)

a12  c  1  2  5  6  8  10 a13  c  1  2  5  6  7  8  9  10 a14  c  2  6  9 a15  c  1  2  5  6  7  9 a16  c  1  2  5  6  8  9 a17  c  2  5  6  7  8  9

j 1

is a positive parameter and

a18  c  2  6  10

a1  c  1  5  7

a19  c  1  2  6  9  10

a 2  c  1  5  8

a 20  c  1  2  5  6  7  10

a 3  c  1  6  9

a 21  c  1  2  5  6  8  10

a 4  c  1  5  6  8  9

a 22  c  2  5  6  7  8  10

a 5  c   2  5   7

a 23  c  2  5  6  7  9  10

a 6  c  1  6  10

a 24  c  2  5  6  8  9  10

a 7  c  1  5  6  9  10

a 25  c  1  2  5  6  7  8  9  10

a 8  c  1  5  6  8  10

Also,

b1  c  1  5  7  8

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b2  c  1  5  6  9 b3  c  1  5  6  7  8  9 b4  c  1  2  5  7 b5  c  1  5  6  10 b6  c  1  6  9  10 b7  c  1  5  6  7  8  10 b8  c  1  5  6  8  9  10 b9  c  2  5  7  8 b10  c  1  2  5  6  8 b11  c  1  2  5  6  7  8  10 b12  c  1  2  5  6  8  9  10 b13  c  1  2  6  9 b14  c  2  5  6  7  9 b15  c  2  5  6  8  9

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Again, the M.T.T.F. of the power plant, in this case, is given by: 

M .T .T .F .W    RSW t dt 0



24 1  1   25 1 1       1      i 1 ai   j 1 b j  1 

   

…(27)

6.3 CASE III: When Failures follow Exponential Distribution Exponential distribution is nothing but a particular case of weibull distribution for   1 and is very useful for practical problem purpose. Therefore, the reliability function for considered power plant at an instant t, in this case, is given by

RSE t    exp . ai t   exp . b j t 25

24

i 1

j 1

(28)

and expression for M.T.T.F., in this case, is given by 

M .T .T .F .E    RSE t dt 0 25 1    i 1  ai

 24  1       j 1  b j

   

…(29)

7. NUMERICAL COMPUTATION

b16  c  1  2  5  6  7  8  9

For a numerical computation, let us consider the values:

b17  c  1  2  6  10

(a)

b18  c  2  6  9  10 b19  c  2  5  6  7  10 b20  c  2  5  6  8  10 b21  c  1  2  5  6  7  8  10 b22  c  1  2  5  6  7  9  10 b23  c  1  2  5  6  8  9  10

i i  1,2  10    0.002,   2

and t  0,1,2   in equation (26); (b) i i  1,2  10    0.002, t  0,1,2   in equation (28).

By using (a) and (b), we compute the table-1 and the corresponding graph has been shown in fig-2. (c) and  2 i i  1,2  10    0, 0.04, 0.09  ,1 in equation (27); (d) i i  1,2  10    0, 0.04, 0.09  ,1 in equation (29).

By using (c) and (d), we compute the table-2 and the corresponding graph has been shown in fig-3.

b24  c  2  5  6  7  8  9  10 and c  3  4

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Table-1

t

R SW t 

R SE t 

0 1 2 3 4 5 6 7 8 9 10

1 0.9979 0.9917 0.9803 0.9628 0.9379 0.9044 0.8614 0.8088 0.7471 0.6778

1 0.9979 0.9959 0.9938 0.9917 0.9895 0.9872 0.9849 0.9826 0.9803 0.9779

M.T.T.F. ----->

IJRET: International Journal of Research in Engineering and Technology

eISSN: 2319-1163 | pISSN: 2321-7308

10 9 8 7 6 5 4 3 2 1 0

M.T.T.F.(W) M.T.T.F.(E)

0 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.81

Reliability-->

Reliability Vs Time 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Failure Rate -----> Fig-3

8. RESULTS AND DISCUSSION

RSE(t) RSW(t)

0 1 2 3 4 5 6 7 8 9 Time--> Fig-2 Table-2

In this study, the author has been obtained various performance measures of considered power plant, by using algebra of logics. The reliability and M.T.T.F. of power plant have been obtained in two different cases, i.e., when failures follow Weibull and exponential time distributions.

The final results have been mentioned in the equations (25) through (29).

A critical examination of fig-2 and 3 shows that performance of considered system decreases with time. In case of Weibull time distribution, performance decreases catastrophically in the beginning but thereafter it decreases in a constant manner. We conclude in last that exponential time distribution gives better results as compared with Weibull time distribution.

λ

M.T.T.F.W 

M.T.T.F.E

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