SME
Process Design · Problem type
Mitigated event frequency (LOPA)
Use this for “A cooling failure happens 0.1 times a year. An alarm (PFD 0.1) and a relief valve (PFD 0.01) protect the reactor. How often does it overpressure?”
Diagram
0.0001 = 1 × 10⁻⁴
10⁻⁴ ≤ 0.0001 < 10⁻³
10⁻⁴ = 0.0001
10⁻³ = 0.001
This is an example. Type your own values to start a new problem.
fₘ (yr⁻¹)
=
f (yr⁻¹)
×
PFD₁
×
PFD₂
Formulas
fₘ
=
f
×
PFD₁
×
PFD₂
1 ×
10
-4
=
0.1
×
0.1
×
0.01
Assumptions
•
LOPA (layer of protection analysis): each protection layer fails on demand with probability PFD.
•
The layers are independent of the initiating cause and of each other, so their PFDs multiply.
•
Frequencies are per year; compare fₘ with the site’s tolerable frequency for that harm.
Step-by-step
WE KNOW
f = 0.1 yr⁻¹, PFD₁ = 0.1, PFD₂ = 0.01
FIND
mitigated frequency (fₘ)
Step 1 · Find mitigated frequency (fₘ)
Use fₘ = f × PFD₁ × PFD₂
Only the share of demands that every layer fails lets the event through, so multiply the frequency by each PFD.
fₘ = 0.1 × 0.1 × 0.01
fₘ = 0.01 × 0.01
fₘ
=
1 ×
10
-4
yr⁻¹
Check
Put the numbers back in. Both sides should match.
1 ×
10
-4
=
0.1
×
0.1
×
0.01
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.