082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

29.15 Reliability engineering 975
k s
=
1
500000 = 2 ×10−6 failuresper hour
k rgs
= 1 = 2 ×10 −4 failuresper hour
5000
So the overall reliabilityof the systemis
R = e −(k ts +k tgs +k s +k rgs )t = e −(1×10−3 +5×10 −4 +2×10 −6 +2×10 −4 )t
= e −1.702×10−3 t
Fort = 28 ×24 hours,R = 0.319.
Fort = 365 ×24hours,R = 3.35 ×10 −7 .Clearlyduringayearlyperiodthesystem
isalmost certain tofail atleastonce.
29.15.2 Parallelsystem
A parallel system is one in which several components are in parallel and all of them
mustfailforthesystemtofail.ThecaseofthreecomponentsisshowninFigure29.24.
The probability of all three components failing in a time period,t, is the product of the
individual probabilities of each component failing, that is (1 −R A
)(1 −R B
)(1 −R C
).
So the overall systemreliabilityis
R=1−(1−R A
)(1−R B
)(1−R C
)
This formulacan begeneralizedtothe case ofncomponentsinparallelquite easily.
A
Input
B
Output
C
Figure29.24
Parallel system.
Engineeringapplication29.12
Reliabilityofprocesscontrolcomputerpowersupplies
A process control computer is supplied by two identical power supplies. If one fails
then the other takes over. TheMTBF of the power supplies is 1000 hours. Calculate
thereliabilityofthepowersupplytothecomputerduringa28dayperiod.Compare
this with the reliabilityifthereisno standbypower supply.
Solution
k = 1 = 0.001 failures per hour
1000
R ps
= e −kt = e −0.001×24×28 = 0.511
➔