1. Reference legislation
2. Input Data
3. Essential Calculations
4. Safety Margin
Make sure there is an adequate safety margin for each parameter, also considering the length and type of cables.
5. Documentation
There probability of failure on demand (PFD) is a key measure in functional safety, specifically defined in EN 61508 standard Part 6 (Ed. 2). Quantifies the probability that a safety function will fail to perform the required action on demand and is typically used to evaluate the reliability of safety systems.
To calculate the PFD for a system according to EN 61508 Part 6, several are needed input parameters key . These parameters help define the reliability of the system and the level of security it can provide.
Key input parameters for PFD calculation according to EN 61508 Part 6, Ed. 2:
Safety Integrity Level (SIL):
SIL 1, SIL 2, SIL 3 or SIL 4
: The required safety integrity level determines the overall reliability and risk reduction required for the safety function. Each SIL has specific PFDs and probability ranges that are acceptable for the system.
The SIL is typically determined based on the evaluation of the risk (or risk matrix ) of the system.
Component failure rate (λ):
λ (Failure Rate):
The failure rate of safety-related components (sensors, actuators, logic solvers, etc.), expressed in
breakdowns per hour
(for example, failures/hour or failures/year).
Failure rates are often based on reliability data provided by manufacturers or specific industry standards such as IEC 61508 , IEC 61511 or FIT (Failures in Time).
Test interval (T):
T (Test interval): The time between tests or inspections of safety functions. This value is important for calculating the PFD, as it takes into account the frequency of testing or maintenance of the system.
The shorter the test interval, the lower the PFD, as testing can help identify failures before a demand occurs.
Mean Time to Repair (MTTR):
MTTR:
the average time to repair a failed component or system.
In the event of a failure, MTTR affects how quickly the system can be restored to a safe state.
Diagnostic Coverage (DC):
DC (Diagnostic Coverage):
the percentage of errors that the diagnostic system is able to detect. It is expressed as a value between 0 and 1 (from 0% to 100%).
Diagnostic coverage helps determine the fraction of faults that can be detected before they cause a dangerous failure. A higher DC means a lower PFD.
Common Cause Failure (CCF):
CCF
: the probability that a failure involving multiple components will occur simultaneously. Common cause failures can occur due to environmental conditions or shared resources.
It is important to account for common cause failures when calculating overall system reliability.
System architecture:
System Configuration
: The system configuration, such as 1oo1 (one over one), 1oo2 (one over two), 2oo2 (two over two), 2oo3 (two over three), etc., determines the number of components required for the system to function properly and influences the PFD calculation.
The architecture will determine whether redundancy is used to improve system reliability (e.g., two sensors in parallel to reduce the probability of failure).
Test run interval:
Test interval (PTI):
This is the scheduled time interval in which the system is fully tested to ensure that it is still functioning properly.
A shorter PTI can reduce the PFD by ensuring that undetected faults are identified before they cause system failure.
PFD Calculation Formula
The PFD for a system can be calculated using the following general formula:
PFD=λ⋅T+(1−DC)⋅MTTR2PFD = \frac{\lambda \cdot T + (1 - DC) \cdot \text{MTTR}}{2}PFD=2λ⋅T+(1−DC)⋅MTTR
Where:
λ
is the failure rate of the component or components,
T
it is the test interval,
A.D
it is the diagnostic coverage,
The MTTR
It is the average time to repair the system.
For more complex architectures, such as systems with redundancy, the PFD is calculated differently based on the specific architecture (e.g., voting systems, 1oo2, 2oo3, etc.).
Typical PFD ranges for SIL
The EN 61508 standard defines the following acceptable PFD ranges for each Safety Integrity Level (SIL):
SIL 1 : PFD = 0.1 to 0.01
SIL 2
: PFD = 0.01 to 0.001
SIL 3
: PFD = 0.001 to 0.0001
SIL 4
: PFD < 0.0001
PFD Calculation Example
Let's assume a system with the following values:
SIL:
2
Failure rate (λ
): 0.0001 failures/hour
Test interval (T
): 6 months (or 4380 hours)
Diagnostic coverage (DC
): 90% (0.9)
Mean Time to Repair (MTTR):
20 hours
Using the simplified formula for PFD:
PFD=(0.0001 failures/hour)⋅(4380 hours)+(1−0.0» 9)⋅(20 hours)2PFD = \frac{(0.0001 \, \text{failures/hour}) \cdot (4380 \, \text{hours}) + (1 - 0.9) \cdot (20 \, \text{hours})}{2}PFD=2(0.0001 failures/hour)⋅(4380hours)+(1−0.9)⋅(20hours) PFD=0.438+0.12=0.5382=0.269PFD = \frac{0.438 + 0.1}{2} = \frac{0.538}{2} = 0.269PFD=20.438+0.1=20.538=0.269
This value would be acceptable for SIL 2 as it falls within the acceptable range between between 0.01 and 0.001 .
Conclusion
To calculate the PFD second EN 61508 Part 6 , it is necessary to carefully evaluate the system failure rates, test intervals, diagnostic coverage, repair times and the overall system architecture. These parameters directly influence the safety integrity and the level of risk reduction provided by the safety function.
To calculate the PFH (Probability of Failure per Hour) According to EN 61508 Part 6, Ed. 2 using the data you provided, we can use a similar approach to the PFD calculation, but applied to obtain PFH values. This calculation focuses on the probability of failure per unit time, which is the key metric for the safety of a system.
Input Data:
Calculating PFH:
The formula for calculating the PFH for a system in accordance with EN 61508 is as follows:
PFH=(λ⋅T)+(1−DC)⋅MTTRTestTime(inhours)PFH = \frac{(λ \cdot T) + (1 - DC) \cdot MTTR}{Test Time (in hours)}PFH=TestTime(inhours)(λ⋅T)+(1−DC)⋅MTTR
Where:
λ is the failure rate per unit time (in FIT)
T is the time interval for the test (in hours)
DC is the diagnostic coverage
MTTR is the Mean Time To Repair in hours
Formula for different systems:
PFH for Single Channel System (1oo1):
PFH1oo1=λ⋅T+(1−DC)⋅MTTR2PFH_{1oo1} = \frac{λ \cdot T + (1 - DC) \cdot MTTR}{2}PFH1oo1=2λ⋅T+(1−DC)⋅MTTR
PFH for 2oo2 System:
PFH2oo2=2⋅PFH1oo1⋅(1−PFH1oo1)PFH_{2oo2} = 2 \cdot PFH_{1oo1} \cdot (1 - PFH_{1oo1})PFH2oo2=2⋅PFH1oo1⋅(1−PFH1oo1)
PFH for 1oo2 System:
PFH1oo2=PFH1oo1⋅(1−PFH1oo1)PFH_{1oo2} = PFH_{1oo1} \cdot (1 - PFH_{1oo1})PFH1oo2=PFH1oo1⋅(1−PFH1oo1)
PFH for 1oo3 System:
PFH1oo3=PFH1oo1⋅(1−PFH1oo1)⋅(1−PFH1oo1)PFH_{1oo3} = PFH_{1oo1} \cdot (1 - PFH_{1oo1}) \cdot (1 - PFH_{1oo1})PFH1oo3=PFH1oo1⋅(1−PFH1oo1)⋅(1−PFH1oo1)
PFH for the 2oo3 System:
PFH2oo3=3⋅PFH1oo1⋅(1−PFH1oo1)⋅(1−PFH1oo1)PFH_{2oo3} = 3 \cdot PFH_{1oo1} \cdot (1 - PFH_{1oo1}) \cdot (1 - PFH_{1oo1})PFH2oo3=3⋅PFH1oo1⋅(1−PFH1oo1)⋅(1−PFH1oo1)
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