212
Chapter 4
= (0.70
Redo C For pressure regulators
with lower extreme
10 -6) x 10=7 x 10-6/hr. Eq. (4.33)
2
7
R4 = exp[-~(8760)]-exp[-2(~06)(8760)l
= 1.8810 - 0.8846
R4 = 0.9965
Rsystem = {0.8778} {1.00} {0.9965}=0.8747 for the top loop
Now since the system is made of two parallel components, Eq. (4.23)
Rsystem = 1 - (1 - Rsy~)(1 - Rsys)
= 1 - (1 - 0.8747)(1 - 0.8747)
Rsystem = 0.9843 yearly overhaul for pump needed for the parallel
setup
(1.57
Qsystem = ~, 1-~] failures
EXAMPLE 4.5. Determine the reliability
of the automotive gear
box Fig. 4.19 noting 3rd gear is used 93% of the time; 2rid gear 3%, 1st
gear 3% and reverse 1%. Find the time to reduce the reliability to 0.90.
The reliability of 3rd, 2rid, 1st, and reverse are each series in components
and the operation of the gear box is a series combination of 3rd, 2nd,
1st and reverse.
The Reliability Model for 3rd Gear
Assuming the driver wishes to operate the car in 3rd gear, maximum speed,
shifts F into the position shown and pushed D to the left Fig. 4.19 so that
the clutch piece C engages with B, in which case P runs at the same speed
as the engine shaft E. This means a series reliability model Eq. 4.24 for
93% of the time shown in Fig. 4.20. In Fig. 4.20 the numbers represent
reliabilities
Rl - A bearing and seal
R2 - A bearing and seal
R3 - A jaw clutch
R4 - Shifting fork
R5 - Left shaft
R 6 - Right shaft
R7 - Housing