t = time (hours) - Note! other units can be used as long as the use is consistent through the calculations.
The failure rate at time t can be expressed as
λ = NF / (No t)
= (No - Ns) / (No t) (2)
NF = No - Ns = number of failing components at time t
Ns = number of live surviving components at time t
No = initial number of live components at time zero
A car manufacturer sells 400000 cars of a certain model in one year. During the the first tree years the owners of 50000 of these cars experience major failures. The failure rate can be calculated as
λ = (50000 cars)/ ((400000 cars) (3 years))
= 0.042 (per year)
The reliability of a model of this car within three years can be calculated as
R = e-(0.042 1/year) (3 year)
= 88 %
The connection between reliability and unreliability:
R + Q = 1 (3)
Q = unreliability. Values between 0 - 1 where value 1 indicates 0% live components and value 0 indicates 100% live components.
(1) and (2) can be used to express unreliability
Q = 1 -e-λt (4)
The unreliability of the car model in the example above within three years can be calculated as
Q = 1 - e-(0.042 1/year) (3 year)
= 12 %
The number of live surviving components in a system at a given time:
Ns = No e-λt (5)
The number of failure dead components in a system at a given time:
Ns = No (1 - e-λt) (6)
Mean time between failures - MTBF:
MTBF = 1 / λ (7)
MTBF = Mean Time Between Failure (hours)
From the example above the failure rate is 0.42 per year. MTBF can be calculated as
MTBF = 1 / (0.042 1/year)
= 23.8 years
Mean Time Between Failure (MTBF) can be determined by rating Total Surviving Hours against Number of Failures as
MTBF = ts / NF (8)
ts = total surviving hours (hours)
Combining (5) with the formulas for reliability and more
R = e-t/m (1b)
Q = 1 -e-t/m (4b)
Ns = No e-t/m (5b)
Nf = No (1 - e-t/m) (6b)
MTBF and MTTF are often used interchangeably.
There are two basic types of reliability systems - series and parallel - and combinations of them.
Reliability of systems in series can be expressed as
Rs = R1 R2 (9)
Rs = system reliability
R1,2 = subsystem reliability
From the example above the reliability of a car over three year is 0.88. If we depends on two cars for a mission - whatever it is - the reliability for our mission will be
Rs = 0.88 0.88
= 77 %
Reliability of systems in parallel can be expressed as
Rs = R1 Q2 + R2 Q1 + R1 R2 (10)
Q1,2 = (1 - R1,2) subsystem unreliability
From the example above the reliability of a car over three year is 0.88. If it is enough with one of our two cars for our mission - the reliability for our mission can be calculated as
Rs = 0.88 (1- 0.88) + 0.88 (1- 0.88) + 0.88 0.88
= 99 %
Note! - two systems in parallel with 99 % reliability vs. one system alone with reliability 88 % shows the power of redundancy.
Risk, reliability and safety in process control systems.
Securing a system and achieve equal wear by alternating pumps in parallel.
Design for Assembly Index indicates how easy it is to assembly a component.
Sizing safety valves according boiler output power in high pressure systems (kW and Btu/hr)
Non-destructive construction testing.
Cylinder volume and compression ratios in piston engines.
Endurance limits and fatigue stress for steels.
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