Availability analysis of parallel transit fuel system in petrol engine under head of line repair

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AVAILABILITY ANALYSIS OF PARALLEL TRANSIT FUEL SYSTEM IN PETROL ENGINE UNDER HEAD OF LINE REPAIR Reena Garg1 1

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

Abstract In this paper, the author has considered a fuel system in petrol engine with one alternative fuel source (like LPG, CNG etc.) keeping in standby arrangement and followed online through a perfect switching device. The whole system can fail due to either its normal working or human error. All the failures follow exponential time distribution where as all repairs follow general time distribution.

Keywords: Fuel system, standby, exponential time distribution, general time distribution.

--------------------------------------------------------------------***-----------------------------------------------------------------1. INTRODUCTION In this paper, the considered fuel system consists of three subsystems, namely A, B and C, connected in series.

Subsystem A is fuel source having two units in standby redundancy. Subsystem B consists of two units (Transfer pump and carburetor) connected in series while subsystem C having N-cylinders [identical] in series and is of j-out-ofN:G nature, fig-1 shows the system configuration. Head-ofline policy is being adopted for repair purpose. The concept of human-error is also incorporated to make the model more compatible. Laplace transform of various state probabilities,

availability and long run availability, some particular cases and profit function for the system are obtained. Fig-2 shows the state-transition diagram and states descriptions have mentioned in table-1.

By using supplementary variables technique and Laplace transform, various reliability parameters have been obtained.

One numerical example together with graphical illustration has also been given at the end to highlight important results of the study.

Table-1: States Description SUBSYSTEM SYSTEM’S SUPPLEMENTARY STATE VARIABLE USED A B C S1 0 0 0 0 Nil S2 0 F 0 F u S3 0 0 (N-j)0 0 Nil S4 0 0 F F v S5 1F 0 (N-j)0 0 Nil S6 1F F (N-j)0 F y S7 0 F (N-j)0 F u S8 1F 0 0 0 Nil S9 1F F 0 F y S10 FH FH FH FH m S11 F 0 0 F n S12 F 0 (N-j)0 F r O: All Operable; F : all Failed; FH : failure due to human-error; rF :r units have failed; rO : r units are operable.

STATE

2. ASSUMPTIONS

3.

The following assumptions have been associated with this model: 1. Initially, all the units are new and the system is in good condition of full efficiency.

2. Repair has given only when the system is in failed condition and a single service channel is available.

4.

After repair, system works like a new one and never damages anything.

1-A unit is a petrol tank while 2-A unit is an alternative fuel source (like CNG, LPG etc.) and the perfect switching device has been used to online the standby unit.

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3. NOTATIONS

repaired up to time s, where K=A, B, C and s=u, v, y, n, r.

The following notations have been used throughout the model: : The probability that at time t, the system PK (t ) is operable and is in Kth state, where K=1, 3, 5, 8. : The probability that at time t, the system PK (s, t ) is in rth failed state and elapsed repair time lies in the interval (s, s + Δ).

K=2, 4, 6, 7, 9, 10, 11, 12 and s=u, v, y, u, y, m, n, r respectively. : Constant failure rate of 1B/2B unit and b1 / b2 note that b1+b2=b (say). a/c : Constant failure rate of any unit of subsystem A/C. H : Human-error rate.  K (s) : The probability that the subsystem K will be repaired in the time interval (s,s+Δ) conditioned that it was not

 h (m)

:

The probability that the system will be repaired from state S10 in time interval (m, m+Δ) conditioned that it was not repaired up to time m.

S i (x )

:

 i ( x) exp . -   i ( x)dx

Mk

:

P

: :

Dk (s )





0





Mean time to repair kth unit   S k (0), for all k.

Time independent probability.

1  S

K



( s) / s , for all k =A,B and C.

Fig-1: System configuration

Fig-2: State-transition diagram

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4. FORMULATION OF THE MATHEMATICAL MODEL Using continuity argument, we obtain the following set of difference-differential equations governing the behavior of the system under consideration:

d   dt  a  b  ( N  j )c  h  P1 (t )    



0

0





0

P2 (u , t )  B (u )du

…(1)

  P10 (m, t )  h (m)dm   P4 ( , t )  c ( )d

    u  t   B (u ) P2 (u , t )  0  

…(2) …(3)

 d   dt  cj  b  a  P3 (t )  ( N  j )cP1 (t )  0 P7 u , t  B (u )du  

…(4)

       t   c ( ) P4 ( , t )  0   …(5)

d   dt  a  b  P5 (t )  aP3 (t )   …(6)

    y  t   A ( y ) P6 ( y, t )  0   …(7)

    u  t   B ( y )  P7 (u , t )  0   …(8)

d   dt  a  b  P8 (t )  aP1 (t )  0 P11 (n, t )  A (n) dn   

…(9)

    y  t   A ( y ) P9 ( y, t )  0   …(10)

     m  t   h (m) P10 (m, t )  0   …(11)

    n  t   A ( n)  P11 ( n, t )  0   …(12)

    r  t   c ( r ) P12 ( r , t )  0   Boundary conditions are:

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P2 (0, t )  bP1 (t ) 





0

P9 ( y, t )  A ( y )dy …(14)

P4 (0, t )  cjP3 (t ) …(15)

P6 (0, t )  bP5 (t ) …(16)

P7 (0, t )  bP3 (t ) …(17)

P9 (0, t )  bP8 (t ) …(18)

P10 (0, t )  hP1 (t ) …(19)

P11 (0, t )  aP8 (t )  



0

P12 (r , t )  c (r )dr …(20)

P12 (0, t )  aP5 (t ) Initial conditions are: …(21)

P1 (0)  1, otherwise zero.

5. SOLUTION OF THE MODEL Taking Laplace transforms of equations (1) through (20) by making use of (21) and then on solving them, we have following Laplace transforms of different state probabilities:

P1 (s) 

1 A( s )

 1  b  a C (s) P 2 (s)   S c ( s ) S A ( s )  D B ( s ) 1  a S A ( s ) A( s)   B( s) s  a  b B( s)  P 3 ( s) 

C ( s) A( s)

…(22) …(23)

…(24)

C ( s) Dc s  A( s )

…(25)

P 5 ( s) 

a C ( s) s  a  b A( s)

…(26)

P 6 ( s) 

ab C ( s) D A ( s) s  a  b A( s)

…(27)

P 4 ( s )  cj

…(28)

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C ( s) DB s  A( s)  a  1 a C ( s) P 8 (s)   S c ( s ) S A s   A( s )  B( s ) s  a  b A( s ) 

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P 7 ( s)  b

…(29)

P 9 (s)  bP8 (s) DA s  P 10 ( s ) 

…(30) …(31)

h Dh s  A( s )

…(32)

  C ( s) a P 11 ( s)  a  P 8 ( s)  S c ( s) D A s  s  a  b A( s)  

…(33)

P12 ( s) 

2

a C ( s) Dc ( s) s  a  b A( s)





where, B(s)  s  b  a 1  S A s 

C (s) 

( N  j )c s  a  cj  b 1  S B ( s )



…(34) …(35)







A(s)  s  a  b  ( N  j )c  h 1  S h (s)  cj S c (s)C(s)  bS B (s)  1  a C (s)  abS A ( s ) S B ( s )  S c ( s) S A ( s)  B( s) s  a  b B( s) 

and

6. ERGODIC BEHAVIOUR OF THE SYSTEM By

making

use

of

Abel‟s

Lemma

viz.,

…(36)

right exists, we have the following time independent state probabilities from equations (22) through (33):

lim sF ( s)  lim F (t )  F ( say) , provided the limit on s 0

t 

P1 

1 A(0)

…(37)

P2 

1  aH   b  a 1   M B  A(0)   a  b 

…(38)

P3 

H A(0)

…(39)

P4 

cjH Mc A(0)

…(40)

P5 

aH 1  a  b A(0)

…(41)

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P6 

abH M a  b A(0) A

…(42)

P7 

bH MB A(0)

…(43)

P8 

a  aH  1  bA(0)  a  b 

…(44)

P9 

a  aH  1 MA  A(0)  a  b 

…(45)

P10 

h Mh A(0)

…(46)

P11 

a2 1  H M A bA(0)

…(47)

a2H M a  bA(0) c N  j c ; and A0   d As  where, H   ds  a  cj   s 0 P12 

…(48)

…(49)

6. SOME PARTICULAR CASES 6.1 When Repairs Follow Exponential Time Distribution: Setting S A ( s ) 

 A / s   A  etc. in equations (22) through (33), we obtained:

P1 ( s ) 

1 E (s)

…(50)

P 2 ( s) 

a A  1  A  1 b  a G( s)  c     1  E ( s)  s   A  F ( s) s  a  b F ( s) s   c s   A  s   B

…(51)

P 3 ( s) 

G ( s) E ( s)

…(52)

G (s) 1 E (s) s   c

…(53)

P 5 ( s) 

a G( s) s  a  b E ( s)

…(54)

P 6 (s) 

ab G ( s ) 1 s  a  b E (s) s   A

…(55)

P 4 ( s )  cj

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P 7 ( s)  b

P 8 ( s) 

G(s) 1 E ( s) s   B

P 10 ( s ) 

…(56)

A  a  1 a G( s)  c    E ( s)  F ( s) s  a  b F ( s) s   c s   A 

P 9 (s)  b P 8 (s)

…(57)

…(58)

1 s  A

…(59)

h 1 E ( s) s   h

 a G( s)  c  1 P11 ( s)  a  P 8 ( s)   s  a  b E (s) s   c  s   A  P 12 ( s ) 

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…(60)

…(61)

a2 G ( s) 1 s  a  b E (s) s   c

where,

F (s)  s  b 

and



as s  A

;

G (s) 

( N  j )c s  a  cj 

E ( s )  s  a  b  ( N  j )c 

bs s  B

c b B hs  cj G( s)  s  h s  c s  B

…(62)

…(63)

 1  c  A ab A  B a G( s)   s   A s   B   F (s) s  a  b F (s) s   c s   A 

6.2 Availability and Cost Function for the System : Now, Laplace transform of availability of petrol engine is given by:

Pup s  

  ( N  j )c 1 a 1   s  a  b  ( N  j )c  h  s  a  cj  b s  a  b 

…(64)

on inverting this, one can obtain the availability at any time „t‟ as under

Pup (t )  Qe a b( N  j ) c  h t  Re a cj b t  Se a b t where

R

Q  1

( N  j )c a  ( N  2 j ) c  h ( N  j )c  h

( N  j )c ( N  2 j )c  h

and S 

…(65)

;

a ( N  j )c  h

…(66) …(67)

Note that Pup 0   1

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Also, Profit function for the system, working up to the time t, is given by …(68)

K t   C1  Pup t dt  C 2 t  C 3 t

0

where,



t

0

Pup t   Q

1  e a b( N  j ) c  h t 1  e a cj b t 1  e a b t R S a  b  ( N  j )c  h a  cj  b ab

…(69)

where, C1 ,C2 and C3 are the revenue, repair and establishment cost per unit time respectively also Q ,R and S are given in the equations (66)and (67).

Availability Vs Time 1

6.3 Steady- State Availability

AVSS

a  Nc a  b   ba  cj a  b  ( N  j )c  h

Where,

H  N  j c / a  cj

(70)

Pup(t)-->

0.8

The steady state availability of the system is given by

0.4 0.2

(71)

0 0

6.4 Numerical Computation

Also t= 0, 1, 2,-----------10 and h=0.001,0.0015, 0.002---------- 0.01.

Table-2 Availability 1 0.976302 0.94806 0.916399 0.882266 0.846451 0.809611 0.772289 0.734929 0.697892 0.661467 0.625883 0.591317 0.557903 0.525739 0.494891

1

2

3

4

5

6

7

8

9 10

t-->

For a numerical computation, let us consider the values: N=5, j=2, a=0.01, b=0.02, c=0.03, h=0.001, C1=Rs.5.00, C2=Re1.00 and C3 =Rs2.00.

Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.6

Fig-3

Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Table-3 Profit Function -2 -0.03425 1.853275 3.650692 5.349202 6.942611 8.426888 9.799798 11.06059 12.2097 13.24855 14.17934 15.00487 15.7284 16.35354 16.88414 17.32424 17.67793 17.94936 18.14267 18.26195

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Profit Function Vs Time 15

K(t)-->

7. CONCLUSIONS The plots of equations (65), (68) and (70) have shown in figs-(3), (4) and (5), respectively.

Fig-3 shows that availability of the system decreases slowly with time initially but there after it decreases approximately in a constant manner.

10 5

Fig-4 shows that profit with the system under consideration.

For t=0 and 1, there is no profit but after this profit increases in constant way.

0 0 -5

1

2

3

4

5

6

7

8

9 10

t--> Fig-4

h 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006 0.0065 0.007 0.0075 0.008 0.0085 0.009 0.0095 0.01

Fig-5 shows long run availability of the system. Its value is 0.283353. Now as we increase the value of h, the long run availability decreases correspondingly.

REFERENCES

Table-4 A VSS 0.283353 0.282187 0.28103 0.279883 0.278746 0.277617 0.276498 0.275387 0.274286 0.273193 0.272109 0.271033 0.269966 0.268908 0.267857 0.266815 0.265781 0.264755 0.263736

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