What is the PFD of a piece of equipment with a failure rate of 0.4 per year and an annual testing schedule?

What is the PFD of a piece of equipment with a failure rate of 0.4 per year and an annual testing schedule?

• A. 0.60
• B. 0.40
• C. 0.16
• D. 0.33

Answer: (D)

To calculate the Probability of Failure on Demand (PFD) for a piece of equipment, we need to consider its failure rate and the testing frequency. The PFD is essentially a measure of the likelihood that a safety system will fail to perform its intended function on demand. In this scenario, the piece of equipment has a failure rate of 0.4 failures per year. Given that there is an annual testing schedule, we can use the formula for PFD, which is derived from the failure rate and the testing interval. The formula is: \[ \text{PFD} = \frac{\lambda}{\lambda + \frac{1}{T}} \] Where: - \( \lambda \) is the failure rate (0.4 per year in this case), - \( T \) is the test interval (1 year, since we are testing annually). Plugging in the values: \[ \text{PFD} = \frac{0.4}{0.4 + \frac{1}{1}} = \frac{0.4}{0.4 + 1} = \frac{0.4}{1.4} \approx 0.2857 \] When approximated, this value is

To calculate the Probability of Failure on Demand (PFD) for a piece of equipment, we need to consider its failure rate and the testing frequency. The PFD is essentially a measure of the likelihood that a safety system will fail to perform its intended function on demand.

In this scenario, the piece of equipment has a failure rate of 0.4 failures per year. Given that there is an annual testing schedule, we can use the formula for PFD, which is derived from the failure rate and the testing interval. The formula is:

[ \text{PFD} = \frac{\lambda}{\lambda + \frac{1}{T}} ]

Where:

• ( \lambda ) is the failure rate (0.4 per year in this case),

• ( T ) is the test interval (1 year, since we are testing annually).

Plugging in the values:

[ \text{PFD} = \frac{0.4}{0.4 + \frac{1}{1}} = \frac{0.4}{0.4 + 1} = \frac{0.4}{1.4} \approx 0.2857 ]

When approximated, this value is