If a control valve in an SIS has a failure rate of once in four years and is tested every three years, the unreliability is nearest to:

If a control valve in an SIS has a failure rate of once in four years and is tested every three years, the unreliability is nearest to:

• A. 3.6 x 10-3
• B. 0.53
• C. 0.3
• D. 0.47

Answer: (B)

To determine the unreliability of the control valve in a Safety Instrumented System (SIS), we first need to understand the concepts of failure rate and test intervals. The failure rate of the valve is given as once every four years, which translates to a failure rate of 0.25 failures per year. Given that the valve is tested every three years, we need to calculate the probability of failure on demand (unreliability) within that test interval. The time until failure can be modeled using an exponential distribution, where the probability of failure in a given time period can be calculated using the formula: \[ P(\text{Failure}) = 1 - e^{-\lambda t} \] Where: - \( \lambda \) is the failure rate (0.25 failures/year), - \( t \) is the time interval (3 years). Calculating this gives: 1. Compute \( \lambda t = 0.25 \times 3 = 0.75 \). 2. Calculate \( P(\text{Failure}) = 1 - e^{-0.75} \). Using \( e^{-0.75} \approx 0.47237 \) (this is a standard value known

To determine the unreliability of the control valve in a Safety Instrumented System (SIS), we first need to understand the concepts of failure rate and test intervals. The failure rate of the valve is given as once every four years, which translates to a failure rate of 0.25 failures per year.

Given that the valve is tested every three years, we need to calculate the probability of failure on demand (unreliability) within that test interval. The time until failure can be modeled using an exponential distribution, where the probability of failure in a given time period can be calculated using the formula:

[

P(\text{Failure}) = 1 - e^{-\lambda t}

]

Where:

• ( \lambda ) is the failure rate (0.25 failures/year),

• ( t ) is the time interval (3 years).

Calculating this gives:

1. Compute ( \lambda t = 0.25 \times 3 = 0.75 ).

2. Calculate ( P(\text{Failure}) = 1 - e^{-0.75} ).

Using ( e^{-0.75} \approx 0.47237 ) (this is a standard value known