EVALUATION OF SOME RELIABILITY PARAMETERS FOR A GAS SEPARATOR PLANT BY BOOLEAN FUNCTION TECHNIQUE S.C Agarwal

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EVALUATION OF SOME RELIABILITY PARAMETERS FOR A GAS SEPARATOR PLANT BY BOOLEAN FUNCTION TECHNIQUE

S.C Agarwal¹ & Mool Pal²

1M.M.(P.G.) College ,Modinagar (U.P.), India.

2TGT (Maths), Kendriya Vidyalaya, Chenani at Kud, Udhampur (J&K), India.

ABSRACT

The authors in this paper have tried to evaluate the reliability and MTTF of a gas separator plant. The considered system consist of four sub system arranged in series and parallel configuration .Boolean function technique have been used to formulate and solve the mathematical model of considered system .All necessary graphical illustrations are given at the end so as to explain the practical utility of the model.

KEY WORD: MTTF And Boolean function technique.

1. INTRODUCTION

In this considered gas separator plant, there are four main sub systems ,named as P,Q,R and S. The subsystem P is a pressure shut down valve and it is responsible to collect input for the separation process. The sub system P is an automatic valve and it work by the pressure of input. The sub system Q consists two unit of pressure logic controller arrange in parallel .These two units control the working of pressure shut down. The subsystem R consists two units arranged in parallel and these unit known as gas separation units. The subsystem R separates the useful gas from the input and exhausted the waste gases. By subsystem S we obtain separated gases.

2. ASSUMPTION

The associate assumptions are as follows.

1. Initially all the equipments are good and operable.

2. The state of each component and of the whole system is either good or fail.

3. The state of all components of the system are statistically independent.

4. The failure time of all components are arbitrary.

5. Supply between any two component of the system is hundred percent reliable.

6. There are no repair facility.

7. The reliability of each component is known in advance.

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RELIABILITY BLOCK DIAGRAM

NOTATIONS

x1 – state of subsystem P

x_2, x₃ – states of two units of subsystem Q x₄, x₅ – states of two units of subsystem R

X1

X2 X3

X4 X5

X6

262 x₆ – state of subsystem S.

x' – negation of x

^ – conjuction

x_i = 0 in bad state

1 in good state ( i =1,2,3,4,5,6)

R_s : Reliability of the system as a whole

R_i : Reliability of component corresponding to system state x_i Q_i : Unreliability of component correspond to system state x_i

Rsw(t)/R_SE (t) : Reliability of the system as a whole when failures follow weibull / exponential time distribution.

3. FORMULATION OF MATHEMATICAL MODEL

The successful operation of the system in terms of logical matrix are expressed as:

F(x ₁,x₂,x₃,x₄,x_5,x₆)= x₁ x₂ x₄ x₆ x₁ x₂ x₅ x₆ x₁ x₃ x₄ x₆

x1 x3 x5 x6 …(1)

4. SOLUTION OF THE MODEL By using application of algebra of logic equation (1) may be written as F(x₁,x₂,x₃,x₄,x₅,x₆)=[x₁.x₆ ^ f (x₂ x₃ x₄ x₅)] … (2) Where f (x₂ x₃ x₄ x₅)= x₂ x₄ x₂ x₅ x₃ x₄ x3 x5 …(3)

Substituting the following in Eqn (3) M₁ = [x₂ x₄ ] …(4)

M₂ = [x₂ x₅ ] …(5)

M₃ = [x₃ x₄ ] …(6)

M₄ = [x₃ x₅ ] …(7)

We obtain f (x₂ x₃ x₄ x₅)= M₁ M₂ M₃ M₄ …(8)

Using orthogonalization algorithm eqn(.8) may be written as f (x₂,x₃, x₄, x₅)= M₁ M'₁ M₂ M'₁ M'₂ M₃ M'₁ M' ₂ M'₃M₄ …(9)

263 Now using algebra of logic, we have

M₁ = [x ₂ x ₄] M'₁ = x'₂

x₂ x'₄ M'₁M₂ = x'₂

x₂ x'₄ ^ [x₂ x₅]

= [x₂ x'₄ x₅] …(10) Similarly we obtain the following

M'₁.M'₂.M₃ = [x'₂ x₃ x₄] … (11) And

M'₁.M'₂.M'₃ M₄ = x'₂ x₃ x'₄ x₅

x₂ x₃ x'₄ x₅ …(12) Using eqn (10) through (12) ,eqn (8) gives

f (x₂ x₃ x₄ x₅)= x₂ x₄ x₂ x'₄ x₅

x'2 x3 x4 x'₂ x₃ x'₄ x₅

x₂ x₃ x'₄ x₅ . . .(13) using eqn (13) in ( 2) we obtain

F(x₁ x₂ x₃ x₄ x₅ x₆) = x₁ x₂ x₄ x₆ x₁ x₂ x'₄ x₅ x₆ x₁ x₂' x₃ x ₄ x₆ x₁ x'₂ x₃ x'₄ x₅ x₆ x₁ x₂ x₃ x'₄ x₅ x₆

Finally, probability of the successful operation (i.e. reliability ) of the system is given by R_s = Pr { F (x₁x₂x₃x₄x₅x₆)}

R_s = R₁ R₆ [R₂R₄+ R₂Q₄R₅+Q₂R₃R₄+Q₂R₃Q₄R₅+R₂R₃Q₄R₅] …(14) R_s = R₁R₆ [R₂R₄+R₂(1-R₄).R₅ +(1-R₂)R₃R₄+(1-R₂)R₃(1-R₄)R₅+R₂R₃(1-R₄).R₅]

R_s = R₁R₆[R₂R₄+R₂R₅+R₃R₄+R₃R₅-R₂R₄R₅-R₂R₃R₄- R₃R₄ R₅ ] …(15) Particular CASES

CASE I : If reliability of each component of the complex system is R, eqn (15) yield

R_S=R⁴[4-3R] ....(16) CASEII;When failure rate follow weibull distribution

Let failure rates of state x_i is λ_i where (i=1,2,3,4,5,6)

Then from eqn (16), reliability of system at an instant t become

R_swt(t) = ∑ exp(– α _it^p ) –∑ exp(– β_jt^p )

i=1 j=1 … (17) α1 = λ1+λ2+λ4+λ6

α2 = λ1+λ2+λ5+λ6 α3 = λ1+λ3+λ4+λ6 α4 = λ₁+λ3+ λ5+λ6

4

264 β1= λ₁+λ2+λ4+ λ₅+λ₆

β2= λ1+λ2+ λ3+λ4+λ6

β3= λ₁+λ3+ λ4+ λ5+λ6

Where α is a possitive parameter.

CASEIII : When failure rates follow exponential distribution

exponential distribution is a particular case of weibull distribution for 𝑝=1,the reliability of the system in this case at instant t given by

…. (18) And the expression for MTTF in this case is

…. (19)

NUMERICAL COMPUTATION FOR RELIABILITY & MTTF

(a) Setting all λi= λ=0.01,i=1to 6, p=2 in eqn (17) and (18) .One may obtain the table 1& fig2.

(b) Setting all λi= λ=0.01,in eqn (19) one may obtain the table 2& fig 3.

Table 1 S.No. t

time

R_SE (t), p=1

Exponential Distribution

R_SW(t),Weibull Distribution

1. 0 1 1

2. 1 0.9990 0.9990

3. 2 0.9980 0.9959

4. 3 0.9970 0.9906

5. 4 0.9959 0.9827

6. 5 0.9949 0.9720

7. 6 0.9938 0.9579

8. 7 0.9928 0.9401

9. 8 0.9917 0.9183

10. 9 0.9906 0.8922

Table 2

S.NO. λ MTTF

1. 0.01 40.0

2. 0.02 20.00

3. 0.03 13.33

4. 0.04 10.00

5. 0.05 08.00

6. 0.06 6.67

7. 0.07 5.71

8. 0.08 5.00

9. 0.09 4.44

10. 0.10 4.00

265 Figure 2

Figure 3

INTER PREPRATION OF THE RESULTS

Table 1 and fig2 shows that reliability function R_SE(t) decreases catastrophically in the beginning but thereafter it decrease constantly .The value of R_SE(t) remains better as compared of R_SE(t).

Table 2 and fig3 show that the values of MTTF with the increase in failure rate. A critical examination of fig 3 ,yields that the value of MTTF decrease rapidly as we make increase in the value of failure rate λ but thereafter in decrease in constant manner.

REFRENCES

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Ind.

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0.8 0.85 0.9 0.95 1 1.05

0 1 2 3 4 5 6 7 8 9

Reliability

Time

Exponential Distribution Weibull Distribution

0 10 20 30 40 50

0 1 2 3 4 5 6 7 8 9

MTTF

FAILURE RATE

Lemda MTTF