SIS ESD Sistems for Process Industries Using IEC 61508 Unit7 SIL Selection

SSlliiddee 11

EIT: E-Cert SS: Unit 7 Instrument Selection

EIT Safety Instrumentation E-Learning

SAFETY INSTRUMENTED SYSTEMS &

EMERGENCY SHUTDOWN SYSTEMS

for Process Industries

using IEC 61511 and IEC 61508

Unit 7: SIL Instrument Selection

www.eit.edu.au

Version for EQO26: 7 November 2012

Presented by Dave Macdonald,

EIT Cape Town South Africa

Contact E-mail: [email protected]

http://www.eit.edu.au/










mailto:[email protected]














Slide 2

Introduction to Chapter 7: Practical selection of

sensors and actuators for safety duties

■ Impact on SIS Reliability,

■ Types of Sensors and Actuators

■ Failure modes and causes

■ Separation, redundancy, diversity, diagnostics

■ Device Selection Issues: What IEC 61511 requires + Common sense

■ Technologies: Safety certified instruments and fieldbus

Knowledge of t he

r ules +

Exper ience…I f

you can get it !

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Key Points about Sensors and Actuators

Slide 3 intelligent instruments

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◆Sensors and Actuators remain the most critical reliability items in an SIS

◆Separation, diversity and redundancy are critical issues.

◆Safety related instruments must have a proven record of performance.

IEC 61508 / 61511 have specific requirements

◆Logic solver intelligence and communications power will help to provide

diagnostic capabilities to assist field device reliability

◆Failure modes and common cause issues are potential problems for












Slide 4

IEC 61511 and other guidance sources

■ Instrument practice for safety systems : well established

■ ISA S 84.01 Appendix B….obsolete standard but still relevant.

■ IEC 61511 specifics defined in clause 11.5 and 11.6 of part 1.

■ Gruhn & Cheddie ISA Textbook; chapter 9

IEC 61511-1 Paragraph 11.5:

Requirements for selection of components and subsystems

■ 11.5.2.1 Components and subsystems selected for use as part of a safety

instrumented system for SIL 1 to SIL 3 applications shall either be in

accordance with IEC 61508-2 and IEC 61508-3, as appropriate, or else they

shall be in accordance with 11.4 and 11.5.3 to 11.5.6, as appropriate

Certified compliant to IEC 61508

Fault tolerance

Prior use

justification

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Sensors and Actuators Dominate Reliability Issues

Slide 5 • PES logic solvers benefit from auto-diagnostics.

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Typical Reliability Table

• The field devices taken together contribute 97% of the PFD for this example.

• The PFD figures for the field devices are affected by environmental conditions

• and maintenance factors.

Table 7.1

Item Fail to

Danger Rate

/ yr.

PFD avg

(3 month proof test)

PFD avg

% of total

Input sensor loop 0.05 0.006 32

SIL 3 Logic Solver PLC 0.0005 3

Output Actuator loop

(Solenoid + valve)

0.1 0.0125 65

Totals 0.019 (SIL 1) 100












Bus connected safety certified instruments

Foundation Field Bus

Profi-safe

ASI-Safety Bus

See Session 5

Slide 6

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Advantages of Analog Transmitters Over Switches

Slide 7

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• Good reliability and accuracy

• Signal present at all times…improved SFF

• Potential for diagnostics, easier to detect faults

• Possible to compare signal with other parameters

• Trending and alarming available

• Multiple set points

• Competitive pricing

• Rationalized spares












Potential Causes of Failures in Sensors

Slide 8

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•Components of the instrument

•Process connection

•Fouling /corrosion/process fluids/clogging

•Wiring

•Environmental: Process/Climate/Electrical

•Specification/range/resolution.

•Response time

•Power supplies

•Intrinsic safety barriers

•Calibration/testing/ left on test/isolated.












Final Control Elements or Actuators

SIS

Logic

Electrical Drive Trip

Interlocks

M

Process Valve Trip

380 v ac

power

Slide 9

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SIS

Logic

Figure 7.4












Slide 10

M

Safet

y

Relay

K1

Relay

K1 Time

Delayed

Reset

Drive

controller

Stop Category 1

Safety Control Category 2

E-Stop

command

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Power

E-Stop operation with VSDlInverter Drive












Slide 11

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· Components of the actuator, positioner, mechanical

failures of springs

Process connection/leaks. Mechanical distortion of

pipes causing stress in valve

Valve internal faults due to : Fouling or corrosion by

process fluids/jamming/sticking/leaking

Wiring to solenoids

Pneumatics/ venting failures

Environmental. Physical impacts/fire/freezing or

icing up.

Solenoid valves sticking or blocking

·

·

·

·

·

Potential Causes of Failures in Final Elements












◆ Sensor contacts closed during normal operation

◆ Tx signals go to trip state upon failure (Normally < 4mA)

◆ Broken wire = trip

◆ Output contacts closed and energized for normal operation

◆ Final trip valves go to trip (safe) position on air failure

◆ Drives go to stop on trip or SIS signal failure

Slide 12

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General Requirements for Fail-safe Operation












For an instrument to qualify for SIL target

Prior Use Build to IEC 61508 HW & SW

Certify to IEC 61508 Smart tx

SIL 1 or 2

SIL 3 requires

assessement and a safety

manual Apply IEC 61511

limitations

Analog or switch

or

And PFD must satisfy SIL target Slide 13

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Sharing of Sensors with BPCS

Slide 14

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Do not share sensors because it:

◆ Violates the principles of independence

◆ Creates a high level of common cause failure

◆ Does not create a separate layer of protection

◆ Does not provide secure maintenance












Boiler Steam

Drum

LT 1

LIC

Feed water

supply

LSL

SIS Logic Solver

Logic

Boiler

Trip

1

Figure 7.5 Snap question: What is wrong with this safety trip

design?

Snap question: Draw a better arrangement Slide 15

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Slide 16

Boiler Steam

Drum

Figure 7.5 cont.

Separate Sensors for Control and Trip: Acceptable

LT 1

Feed water

supply

LIC

1

SIS Logic Solver

Logic

Boiler

Trip

LT

2

LSL

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Slide 17

AND

FW Fails

OR

FW Fails LT-1 Fails

high, LIC-1

causes low

level

0.2 / yr.

0.1 / yr.

PFD = 0.1/2 X 0.5

= 0.025

Trip fails on demand from

FW failure

PFD = 0.1/2 X 0.5

= 0.025

0.2 / yr.

0.005 / yr.

0.1 / yr.

Fault Tree Analysis for Boiler Low Level Trip

Shared Sensor Separate Sensor

Boiler Damage Boiler Damage

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OR

LT-1 Fails

high-No Trip

LIC causes

low level

AND

Low level

0.3 / yr.

LT-2 Fails high

Trip fails on

demand

0.0075 / yr.

Low level and NO TRIP

FW Fails and

No Trip

0.105 / yr.

Low level and NO TRIP

Figure 7.6












Slide 18

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Separation Rules: Field Sensors IEC 61511 part 2 : 11.2.4

•Sharing of sensor between SIS and BPCS only allowed

if safety integrity targets can be met. This would require

sensor diagnostics and is only likely to be possible for

SIL 1

•Separate sensor is allowed to be copied to BPCS via

isolator

•SIL 2, 3 and 4 normally require separate sensors with

redundancy

•SIL 3 and 4 normally require separation and diverse

redundancy












Slide 19

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Separation Rules: Final Elements IEC 61511 part 2 : 11.2.4

•A single valve may be used for both BPCS and SIS but

is not recommended if valve failure places a demand on

the SIS.

•Normally shared valve can only be used if: Diagnostic

coverage and reaction time are sufficient to meet

safety integrity requirements

• Recommendations for a single valve application

•SIL 2 and SIL 3 normally require identical or diverse

separation. Diversity not always desireble












Slide 20

Arrangement for Tripping of Shared Control Valve: SIL 1

SIS

BPCS

FY

FV

A/S

Check hazard demands due to valve

Positioner

Solenoid valve

direct acting,

direct mounted.

De-energise to

vent actuator.

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Figure 7.7











EIT: E-Cert SS: Unit 7

SIS BPCS

Instrument SelectFioingure 7.8

Slide 21 Check hazard demands due to valve

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Diverse Separation of Control and Shutdown Valves SIL 2 and SIL 3

A/S

FY












Sensor Diagnostics

Slide 22

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♦Do not confuse with proof testing

♦Compare trip transmitter value with related

variables. Not often practicable

♦Use safety transmitters… if available

♦Use Smart transmitters with diagnostic alarm

…but see next












Valve Diagnostics

Slide 23

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Assurance that a trip valve will respond correctly when needed

• Freedom of movement, full travel

• Correct venting of actuator

• Correct rate of response

•Absence of sticking

• Trip signals and solenoid all working












Methods for Valve Diagnostics

Slide 24

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• On–line trip testing

• Discrepancy alarm

• Position feedback – response testing

• Partial closure testing – manual or automatic

• Smart positioners – certified safety positioner












IEC Architectural Constraints as per IEC 61508

Slide 25

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◆IEC 61508 places an upper limit on the SIL that can be

claimed for any safety function on the basis of the fault

tolerance of the subsystems that it uses.

♦Limit is a function of

♦the hw fault tolerance

♦the safe failure fraction

♦the degree of confidence in the behaviour under fault

conditions

Details in IEC 61508 part 2












IEC 61508 Classification of Equipment

Slide 26

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◆IEC Defines two types of equipment for use in Safety

Systems:

♦Type A: Simple Devices: Non PES. E.g Limit switch, level

float switch, analogue circuits.

♦Type B: Complex Devices: Including PES. E.G Smart

transmitters. Digital communications, processor based systems.

Fault tolerance rating of B is less than A except under certain

conditions












IEC 61511-1 Table 6: Minimum hardware fault tolerance of

sensors, final elements and non PES logic

SIL Minimum HW Fault Tolerance

1 0

2 1

3 2

4 Special requirements: See IEC 61508

The following summarized conditions apply for SIL 1,2 and 3 :

Increase FT by 1 if instrument does not have fail safe characteristics

Decrease FT by 1 if instrument meets 4 conditions.

•Predominately fail safe

•Prior Use ( Proven in use)

•Limited device adjustment (process parameters only)

•Password protected

Slide 27 Alternatively tables 2 and 3 of IEC 61508 may be applied with an assessment

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Example for Level Switch: Extract from device’s safety manual

Slide 28

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Example for Level Switch: Extract from safety manual

Slide 29

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Example for Level Switch: Extract from safety manual

Slide 30

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Redundancy Options

Slide 31

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Table 7.4

Sensor or Actuator

Configuration.

Selection

1oo1 Use if both PFD and FT and nuisance trip

targets are met.

1oo2 2 Sensors installed, 1 required to trip. PFD

value improved, nuisance trip rate doubled.

2oo3 3 Sensors installed, 2 required to trip. PFD

improved over 1oo1, nuisance trip rate

dramatically reduced.












Common Cause Failures in Sensors

Slide 32

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♦Wrong specification

♦Hardware or circuit design errors

♦Environmental stress

♦Shared process connections

♦Wrong maintenance procedures

♦Incorrect calibrators












Comments on Redundancy in Sensors

Be careful to analyze

for common cause

faults

e.g Try to avoid this

PT

1B PT

1A

SIS

Figure 7.10

Slide 33

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Comments on Diverse Redundancy in Sensors

Where measurement is

the problem use diverse

redundancy.

e.g. Steam or Ammonia

overpressure protection

TT

01

PT

01

SIS

Figure 7.11

Slide 34

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Requirements for Device to be “Proven–in-use”

• Evidence that the instrument is suitable for SIS

• Consider manufacturer’s QA systems

• PES devices need extra validation

• Performance record in a similar profile

• Adequate documentation

• Volume of experience, > 1 yr exposure per case.

Collect t he r ecor ds

of ever y maint enance

event per

inst r ument .

Slide 35

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The approved safety instrument list

• Each instrument that is suitable for SIS

• Update and monitor the list regularly

• Add instruments only when the data is adequate

• Remove instruments from the list when they let you down

Key j ob f or

maint enance

t eam

• Adequate details: Include the process application Slide 36

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Additional requirements for smart transmitters

and actuators:

Details in IEC 61511 11.5.4 for devices with

“Fixed Programming Languages” (FPLs)

Extra for SIL 3

•Formal assessment…low probability of failure in planned

application.

• Appropriate standards used in build

• Consider manufacturer’s QA systems

• Must have a safety manual Slide 37

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Hart Transmitter With Diagnostic Input

Smart

Transmitter

4-20 mA + FSK Data

AI

Hart

Interface Status Alarm

DI

SIS Logic Solver

Hand Held

Programmer

Slide 38 FSK = Frequency Shift Keyed

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Figure 7.12












Example of a Safety Critical Transmitter Figure 7.14

Slide 39

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Benefits of a Safety Certified Transmitter:

Slide 40

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• Internal diagnostics with high coverage factor

• Very low PFDavg values. Saves on proof testing etc.

• Certified for single use in SIL 2 (instead of dual channel)

• Certified for dual redundant use in SIL 3 (instead of 1oo3)

• End user verification is simplified












Importance of the Safety Manual

Slide 41

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The safety manual presents all the essential information and set

up conditions that must be followed to allow the instrument to

be validated for any given application.

The manual also supplies the failure rates summary and

expected PFDavg

Compliance to safety manual requirements must be

demonstrated in the validation phase.

See examples of safety manuals and FMEDA reports












Importance of the Safety Certificate

Slide 42

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The safety certificate is issued by the testing body to clearly define what

products have been tested and what standards and limitations have been

applied in the evaluation.

The safety certificate is an essential document for the validation phase.

See examples of Safety Certificates: 3051C and Rex Radar

Testing Authorities include :

TUV Rheinland

Exida.com

Any recognized testing body that can show competency in the SIS field.

Note : Exida specializes in certifying instruments claiming “prior use”

qualification. Reports supply SFF and failure rate data with declaration of fault

tolerance requirements relevant to IEC 61511. See examples.












Field Devices Summary

Slide 43

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Instruments must be well proven for safety with an assessment

report or Certified SIL capable to IEC 61508.

• Intelligent instruments treated as PES

• Separation, Redundancy, Diversity, Diagnostics

• Diagnostic Coverage via Smarts or Logic Solver

• Bus technology established and growing.











Slide 44

EIT EQO26: Unit 8 Reliability Analysis

EIT Safety Instrumentation E-Learning

SAFETY INSTRUMENTED SYSTEMS &

EMERGENCY SHUTDOWN SYSTEMS

for Process Industries

using IEC 61511 and IEC 61508

Unit 8: Reliability Analysis

www.eit.edu.au Slide 44

Version for EQO26: 7 November 2012

Presented by Dave Macdonald,

EIT Cape Town South Africa

Contact E-mail: [email protected]

























The task of measuring or evaluating the SIS design

for its overall safety integrity

• Reasons and objectives

• Resolving the SIS into reliability block diagrams

• Identification of formulae

• Trial calculation examples

• Calculation software tools


Introduction to Chapter 8:

Reliability Analysis of the SIS












IEC 61511 requires reliability analysis be done for each SIF to show that SIL target and RRF can be achieved. Why?


• Because it tells everyone what RRF can be expected from each

individual safety function.

• It confirms the basis of the design and the chosen proof test

interval

• Compares the calculated RRF for your design with the target to

show you can achieve the target.

• To predict the accident rate: H events/yr = Demand Rate (D) x

PFDavg or H = D/ RRF












Terminology


RRF Risk Reduction Factor ( e.g. 200)

SIL Safety Integrity Level ( depends on RRF)

(SIL Tables)

D Demand rate on Safety Function. ( How often the SIF is

demanded to respond to a hazard condition)

H Hazardous event rate ( also called accident rate )

( e.g. 0.1/yr = 1 in 10 years)

PFDavg Average probability of failure on demand of the SIF












Terminology


MTTFd Mean time to fail dangerously ( = 1/Zd)

MTTFs Mean time to fail safe (or spurious) ( = 1/Zs)

MTTRd Mean time to detect and repair a dangerous fault

Ti Time interval between proof tests

Zdd Failure rate for dangerous detectable faults

Zdu Failure rate for dangerous undetectable faults (requires

proof testing)

Zsd Safe revealed failure rate ( causes spurious trip or loss of

affected safety channel)












Risk Reduction Factor and PFDavg


(PFDavg = average probability of failure on demand,)

PFDavg is a function of:

1. Failure rate per hour for undetected faults : Ldu

2. Test interval: Ti

3. Redundancy (1oo1, 1oo2, 2oo3, etc)

Compare PFDavg with the target PFDavg for the SIL range we need.

RRF = 1

PFDavg












1 Because it can tell you the accident event rate

H = Demand Rate x PFDavg

2 Because it helps you decide the SIL of your design

PFDavg defines the SIL range for the design

(in terms of resistance to random hardware failures

Snap Question: Why is PFD so useful to know?












EIT EQO26: Unit 8 Reliability A nalysis

occurs

Operating but

not protected

Mission time

State of Process

Operating

safely

Hazardous condition

occurs (Demand)

Reportable

accident

1 yr 2 yr

Failure scenario for an Untested SIF

Unrevealed Dangerous fault

occurs












EIT EQO26: Unit 8 Reliability Analysis S

tate

of

Pro

cess

Operating

safely

Operating but not

protected

Hazardous condition

Occurs (Demand)

Accident

prevented

Proof test reveals

fault

Fault

repaired

Low Demand Mode: Proof Tested SIF repaired before demand

Unrevealed Dangerous

fault occurs

Proof test

Mission time 0.5 yr 1 yr












EIT EQO26: Unit 8 Reliability Analysis S

tate

of

Pro

cess

Operating

safely

Operating but not

protected

Demand occurs

before next proof

test

Failure (to respond)

on Demand

Low Demand Mode: Proof tested SIF but failure on demand

Unrevealed Dangerous

fault occurs

Reportable

accident

occurs

Proof test

Mission time 0.5 yr 1 yr













State of Process

Detectable Dangerous

fault occurs

Operating safely

Diagnostic test

reveals fault

Proof test for

undetected

faults

Diagnostic + Proof Tested SIF

Accident

prevented

PFDavg = MTTD&R x Fail danger rate

Fault

detected &

repaired

Mission time Diagnostic test

typically100

wwtwim.eeits./eddauy.au

1 yr 2 yr

Slide 54


Low Demand Mode versus High Demand Mode

• Low demand mode applies when the demand on the SIS is equal to

or less than once per year. ( IEC 61511) . Alternatively no more than

two demands per proof test interval.

• Low demand calculations use PFDavg.

• Hazard event rate H = D x PFDavg

• High demand mode applies when the demand on the SIS is more

than once per year. ( IEC 61511) . Alternatively more than two

demands per proof test interval.

• High demand mode calculations use PFH probability of dangerous

failure per hour.

• Hazard event rate H = PFH

(High demand also known as continuous mode)













Low Demand Mode Application

Pressure relief

trip (SIS)

Pressure surge

once per year

(D)

Accident occurs if

dangerous fault

undetected before the

surge occurs


Accident rate H = D x PFDavg

Provided Test interval is shorter than 1 year or

diagnostics detect faults quickly

Example: If PFDavg = 0.05 and D= 1 : H = 0.05/yr












High demand Mode Application

Electronic

Braking Controls

(SIS)

Brake applied

100 times per

day

Accident occurs as

soon as brake circuit

fails


Accident rate = Probability of failure/hr of the EBC

= Failure rate per hour of the SIS

Example: If PFH = 0.0001/hr H = 0.0001/hr of service

If machine used for 5000 hrs /yr accident rate = 0.5/yr.












Design Iteration for Target PFD in Low Demand Mode

Set Target PFD

Evaluate Solution PFD

Revise Design

No

Yes

Proceed to Detail Design

Acceptable

SRS defines the Risk Reduction Factor

PFD = 1/RRF

Calculated PFD < Target PFD?













Elements and terms in the SIS model

(SIS) Hazard

Demand Rate D H

Protective System

Hazard

Event Rate

PFD avg. = H/D = 1/(Risk Reduction Factor)

SIL3

SIL2

SIL1

Sensor Logic Actuator D H

PFD1 PFD2 PFD3

Overall PFD = PFD1 + PFD2 + PFD3













Single Channel Basic calculation of PFD

How is this formula obtained ?

Zdu

If the fail to danger rate is Zd and proof test interval is Ti


PFDavg = Zdu x Ti/2 (failure rate/yr x mean time to detect )

Example Fail to danger rate = 0.05 per year, Ti = 1 year

PFDavg = 0.05 x ½ = 0.025. ( SIL 1)












Hazard Rate v Demand Rate showing low and high demand modes

D x T<< 1

Accident Rate H = PFH of SIS

Continuous mode

Demand rate D

Hazard

Event

Rate H

H = Ld

D x T> 1

Accident Rate

H = Fail rate Zd

H = Ld ( 1–e - DTi/ 2)


Demand mode

Accident Rate H = Demand

Rate (D) x PFD avg of SIS












Effect of Manual Proof Testing …. leading to average probability of failure on demand:

Time t

p(t)

Probability of

being failed when

demand occurs.

1

0

p(t) = Ld .t

Ti 2Ti

PFDavg = Ld .Ti/ 2

Proof test action

Average

value













SIS Failure Modes

Overt Failures

Spurious Trip Rate

λS = 1/MTBFsp

Loss of Production

Detectable

by Self

Diagnostics

Undetectable

except by manual

proof testing

Trips plant unless

2oo3 or 2oo2 voting

Covert Failures

Dangerous Failure Rate

λD = 1/MTTFD

λD

λDU λDD

ZDU = (1 –C) ZD


ZDD = C ZD

λS + λDD

C= Coverage











EIT EQO26: Unit 8 Reliability Analysis Example: Find the Safe and Dangerous Failure Modes

SIS H igh Level T rip

Logic Solver

LT

1

PSV

LC

1

I/P

FC

Fluid

Feed FC

LT

2

AS


Assume out of range detection provided (forcing a trip)

Fail Modes/yr Device Lsp Ldu Ldd

Bottom Blocked : 0.1 . Top leaks 0.2 LE connection

Runs low: 0.05. Runs high : 0.02 LT electronics

Breaks: 0.01 Shorts across LT: 0.1 Cable

Lost power: 0.02 Power

Totals for sensor sub system:











Overt Failures

Spurious Trip Rate

λS = 1/MTBFsp

Loss of Production

Detectable by

Self

Diagnostics

Detectable by

manual proof

testing

Trips plant unless

2oo3 or 2oo2 voting

Covert Failures


λD = 1/MTTFD

λD

ZDU = (1 –C) ZD

λS + λDD

C= Coverage

λDD = C λD

PFD1 = λDD x (MTTR) PFD2 = λDU x (Ti/2)


1oo1 SIS Formulae

Single Channel SIS Fail Rates


SP Trip Rate = λs + λDD













Overt Failures

Spurious Trip Rate

λS = 1/MTBFsp

Loss of Production

Detectable by

Self

Diagnostics

Detectable by

manual proof

testing

Trips plant unless

2oo3 or 2oo2 voting

Covert Failures


λD = 1/MTTFD

λD

ZDU = (1 –C) ZD

C= Coverage

λDD = C λD

SP Trip Rate = 2 ( λs + λDD)

1oo2 SIS Formulae

PFD2 =((λD U .Ti)2)/3 PFD1 =2(λDD)2( MTTR)2














Overt Failures

Spurious Trip Rate

λS = 1/MTBFsp

Loss of Production

Detectable by

Self

Diagnostics

Detectable by

manual proof

testing

Trips plant unless

2oo3 or 2oo2 voting

Covert Failures


λ = 1/MTTF

λD

D D

ZDU = (1 –C) ZD

λS + λDD

C= Coverage

λDD = C λD

Formula sets

Formula set 2

in Fig 8.6

Formula set 3

in Fig 8.6

Formula set 1

in Fig 8.6













Overt Failures

Spurious Trip Rate

λs = 1/MTBFsp

By Self

Diagnostics

By Manual

Proof testing

λs 1oo1

2λs 1oo2

2(λs)2(MTTR) 2oo2

λD U (Ti/2) λD D (MTTR)

((λD U .Ti)2)/3 2(λDD)2( MTTR)2

λD U .Ti 2 λD D (MTTR)

6(λD D)2 (MTTR)2 2oo3 6(λs)2(MTTR)

Detectable

Spurious trip rate PFD due to diagnostics

(if detected but not tripped)

Multi-channel Formula Sets for PFD and λs (excluding

common mode failures ) Covert Failures


λd = 1/MTTF

PFD due to proof test

Detectable

Formula set 1 Formula set 2 Formula set 3

λD D = DC. λD λD U = (1-DC) λD

Voting

((λD U .Ti)2)

Figure 8.6














Sources of Reliability Data

http://www.sintef.no/Projectweb/PDS-Main-Page/PDS-Handbooks/

Sintef: http://www.sintefbok.no/Product.aspx?sectionId=65&productId=559&categoryId=10

Also see:

Reliability Handbook 1. exida.com

2. Manufacturers’ Safety manuals for

specific SIL certified instruments

3. Faradip 3 Database

4. exida.com: Safety Automation

Equipment List ..Functional Safety

Assessment Reports

http://www.exida.com/index.php/resour

ces/sael/

















































http://www.sintefbok.no/Product.aspx?sectionId=65&productId=559&categoryId=10





































































Dual Channel Basic calculation of PFD Note: Zdd omitted for clarity


Zdu

Zdu

If the fail to danger rate is Zdu and proof test interval is Ti.

PFDavg = (Zdu xTi)2 /3

Example: If fail to danger rate = 0.05 per year, Ti = 1 year

PFDavg = (0.05 x 1)2 / 3 = 0.00083 ( SIL 3)

But this ignores common cause and is unrealistic











EIT EQO26: Unit 8 Reliability Analysis Beta Factor: Common Cause Failures in redundant SIS channels

Unit Failures

(1-β) λd

(1-β) λd

(1-β) λd

Common Cause

Failures

β λd

Example:

2oo3 sensor with

common cause

failures













Formulae Sets with Common Cause Factor included













Dual Channel Basic calculation of PFD inc Common Cause 5%

Note: Zdd omitted for clarity


(1-β) λdu

If the fail to danger rate is Zd and proof test interval is Ti.

PFDavg = ((1-β) λdu xTi)2 /3 + β λdu xTi/2

Example Fail to danger rate = 0.05 per year, Ti = 1 year Beta = 5%

PFDavg = (0.95 x 0.05 x 1)2 / 3 + (0.05 x 0.05 x ½) = 0.002 ( SIL 2)

β λdu (1-β) λdu












2oo3 Channel Basic calculation of PFD inc Common Cause 5%

(1-β) λd

(1-β) λd

If the fail to danger rate is Zd and proof test interval is Ti.

PFDavg = ((1-β) λdu xTi)2 + β λdu xTi/2

Example Fail to danger rate = 0.05 per year, Ti = 1 year Beta = 5%

PFDavg = (0.95 x 0.05 x 1)2 + (0.05 x 0.05 x ½) = 0.0035 ( SIL 2)

β λd (1-β) λd













Formulae Sets with Common Cause Factor included













Calculation Table for PFDavg

Worked example for 1oo1

Formula for calculating PFDavg for 1oo1

PFDavg = (LDU xTi/2) + (LDD x MTTR)

Failures per year


Parameter Value Notes

LDU 0.0500 Dangerous undetected failure rate for one channel

LDD 0.1000 Dangerous detected failure rate for one channel

Ti in yrs 1.0000 Proof test interval

MTTR in yrs 0.0027 Mean time to detect and repair a detectable fault

(LDU xTi/2) 2.50E-02 Undetected portion

(LDD x MTTR) 2.74E-04 Detected portion

PFD for 1oo1 subsystem 2.53E-02 SIL Table: SIL 1












Calculation Table for PFDavg



PFDavg = (LDU xTi/2) + (LDD x MTTR)

Failures per hour



LDU 5.71E-06 Dangerous undetected failure rate for one channel

LDD 1.14 E-05 Dangerous detected failure rate for one channel

Ti in hrs 8760 Proof test interval

MTTR in hrs 24 Mean time to detect and repair a detectable fault

(LDU xTi/2) 2.50E-02 Undetected portion

(LDD x MTTR) 2.74E-04 Detected portion

PFD for 1oo1 subsystem 2.53E-02 SIL Table: SIL 1












Formatted Calculation Table for PFDavg

Worked example for 1oo2 (1-β) λd


PFDavg = (1/3)*((1-þ)LDU xTi)2 + 2((1-þ)LDD x MTTR)2 +þ(LDU xTi/2)+þ(LDD)x MTTR


Failures per year

β λd (1-β) λd

Safecalc: LD = 1.71

% safe =0 C=66%




þ 0.1000 Common cause factor for dangerous and safe failures



(1/3)*((1-þ)LDU xTi)2 6.75E-04 Undetected Voting portion

2((1-þ)LDD2 x MTTR2) 1.18E-07 Detected voting portion

þ(LDU xTi/2) 2.50E-03 Undetected Common portion

þ(LDD)x MTTR 2.70E-05 Detected common portion

PFD for 1oo2 subsystem 3.20E-03












Formatted Calculation Tables for PFDavg



PFDavg = ((1-þ)LDU xTi)2 + 6((1-þ)LDD x MTTR)2 +þ(LDU xTi/2)+þ(LDD)x MTTR

Failures per year

β λd (1-β) λd

(1-β) λd

(1-β) λd





þ 0.1000 Common cause factor for dangerous and safe failures



(1-þ)LDU xTi)2 2.03E-03 Undetected Voting portion

6((1-þ)LDD x MTTR)2 3.54E-07 Detected voting portion

þ(LDU xTi/2) 2.50E-03 Undetected Common portion

þ(LDD)x MTTR 2.70E-05 Detected common portion

PFD for 2oo3 subsystem 4.55E-03












SIS Analysis Model Example

Proof

Testing

Auto

Diagnostics

Proof

Testing


Failure Rates: Z

or MTTF

0.01 0.005 0.01

Overall PFD avg. = 0.025

= 2.5 E-2

Qualifies for SIL 1 (E-1 to E-2)

Apply

Testing or

Diagnostics

PFD averages:

Apply

calculation

+ +


d1=0.2 Zd2=0.02

Zd3=0.1

5yrs 50yrs 10yrs












SIS Analysis: Step 1

(SIS) Hazard

Demand Rate D H

Protective System

Hazard

Event Rate


SIL 2 SIL 1 SIL 1

SIL 1













SIS Analysis: Step 2, identify channels in each stage


Sensor


Logic

Actuator D H

Senso

r

1oo2D

Actuator

1oo2

D H

Example:Dual channel sensors and actuators, single channel logic

1oo1D












SIS Analysis: Step 3, expand details for each single channel

Sensor

Logic

Sensor

1oo2D

1oo1D


Process

Connection Transmitter

Cable and

Power

Expand detail of sensor sub system and apply fail rates for each item











EIT EQO26: Unit 8 Reliability Analysis SIS Analysis: Step 4: Decide λdu, λdd and λs for the elements Step 5: Enter the values to table and totalize

Process

Connection

λDU1

λDD1

λSD1


Transmitter Cable and

Power

λDU3

λDD3

λSD3

λDU2

λDD2

λSD2

Subsystem

Element

Device LSD/hr LSU/hr LDD/hr LDU/hr

1 Process connection 1.14E-05 0.00E+00 5.71E-06 3.42E-06

2 Transmitter 1.14E-05 0.00E+00 5.71E-06 5.71E-07

3 Cable and Power 1.14E-05 0.00E+00 5.71E-06 3.42E-06

4

5

Subsystem totals 3.42E-05 0.00E+00 1.71E-05 7.42E-06












SIS Analysis: Step 6, find the PFDavg for the 1oo2 subsystem

β = common cause failure fraction

1oo2 Failures common to

Ch1 and Ch2 sensors Logic

1oo1 β λd

Redundant section:

PFDavg =

2((1-β).λdd)2 . (MTTR)2

+ ((1-β) .λdu .Ti)2)/3

Common cause section

PFDavg =

β .λdd (MTTR)

+ β .λdu . Ti/2)

+

(1-β) λd

(1-β) λd

= PFDavg

Break out the common cause failure fraction for the redundant channels and calculate

PFD for each portion and add them together













SIS Analysis: Step 7, repeat steps 3 to 6 for each stage

Sensor

Logic

Actuator

Senso

r 1oo2

Actuator

1oo2

Example: Dual channel sensors and actuators, single channel logic

1oo1

PFDavg

for sensors +

PFDavg for

logic solver +

PFDavg

for actuators













SIS Analysis: Example

Example: Dual channel sensors and actuators, single channel logic. 1yr test

.045

0.05

.09

.045 .09

1oo2

1oo1D

λDD = 0.0475 1oo2

Dual Sensors PFD

= .00075 +.00125

= .002

Logic solver PFD

= .00013 +.00125

= .00138

Dual Actuators PFD

= .005 + .0027

= .0077

.0025 .01

SIS PFD = .002 + .0014 +.0077

= . 0111 or 1.11 E-2 = SIL 1


þ = 5% þ = 10%

λDU = 0.0025

C = 95%

λDU = 0.05 λDU = 0.1












SIS Analysis: Example using the EIT Calculator


me: EIT GP SIL Calculator .xls Data Input Table for Sensor Subsystem File na

Proof Test Interval in Hrs (Ti) 8760

Common cause factor (B)% 5%

Mean Time To Test & Repair (Hrs) (MTTR) 24

Subsystem

Element

Device LSD/hr LSU/hr LDD/hr LDU/hr

1 Sensor all components 1.14E-05 0.00E+00 0.00E+00 5.71E-06

2

3

4

5

Subsystem totals 1.14E-05 0.00E+00 0.00E+00 5.71E-06

Calculation results for Sensing

Safe Failure Fraction 66.7%

Diagnostic coverage 0.0%

PFDavg for 1001 2.50E-02














IEC Table of PFDs relevant to Figure 8.16













Honeywell Safecalc example relevant to fig 8.16













SIS Analysis: Example Calculation for Spurious Trip

Example:Dual channel sensors and actuators, single channel logic

Sensor MTTF = 5 years, 75% safe failure fraction. C=0%, β = 10%, Ti = 0.5 yrs, MTTR = 8hrs

Logic MTTF = 10 years, 50% safe failure fraction. C= 95%, β = 10%, Ti = 1 yr

auto diagnostics test interval = 2 secs, MTTR = 24hrs

Actuator MTTF = 2 years, 80 % safe failure fraction. C= 0%, β = 10%, Ti = 0.25 yrs, MTTR =

24hrs

Sensor: single channel λs = 1/5 x .75 = .15/yr

Logic: single channel λs = 1/10 x .5 = .05

Actuator: single channel λs = 1/2 x .8 = .4/yr


λdd = (C x λd ) =95% x 0.05 = .0475/yr











EIT EQO26: Unit 8 Reliability Analysis SIS Analysis: Example Calculation for Spurious Trip

Example :Dual channel sensors and actuators, single channel logic


Spurious Trip for 1oo1

ST = LS + LDD Logic solver 1oo1

Parameter Sensor Logic Actuator Notes

LS 0.05 Fail safe rate

LDD 0.0475 DD rate added due to 95 coverag

Total for 1oo1 subsystem 0.0975 Spurious trip rate per yr

Spurious Trip for 1oo2

ST = 2x(1-B) (LS + LDD) +B(LS + LDD) Actuators: 1oo2

Parameter Sensor Logic Actuator Notes

LS 0.15 0 0.4 Fail safe rate

LDD 0 0 0 DD rate added due to S

Beta 0.1 0 0.1

2x(1-B) (LS + LDD) 0.27 0 0.72 1oo2 portion

B(LS + LDD) 0.015 0 0.04 Common portion

Total for 1oo2 subsystem 0.285 0 0.76 Spurious trip rate per yr

Overall Spurious Trip Rate

1.1425 per yr












SIS Analysis: Example, Spurious Trip Rate

Example: Dual channel sensors and actuators, single channel logic

.36

..0135

.05

.0135 .36

1oo2

1oo1

1oo2

Dual Sensors Spurious

= .28 trips per yr

Logic solver

.097 trips per

yr

Dual Actuators PFD

= (2x .36) + (1x.04)

= .76 trips per yr

.04

Spurious trip rate = ..28 + .097 +.76

= 1.14 trips per year

.015













Reducing Spurious Trip Rate

.135

.015

.135

.135

2oo3 Sensors Spurious

= 6x λs2 (MTTR)+ β λs = (6 x .1352x 8/8760) + .015

= .0001 + .015

. 015 trips per yr

2oo3

.15

1oo2

Dual Sensors Spurious

= 2 x .15

= .30 trips per yr

From 0.3 per year to 0.015/yr

If 1 trip costs AUD 50 000 the annual saving is

What? ……………………………….


.15

Design Version A

Design Version B












Outcomes of a Reliability Study


• Show whether or not the SIS will satisfy the SIL target

• Overall SIS Probability of Failure on Demand (PFDavg)

• PFDavgs for each section of the SIS

• Show benefits of redundancy or voting schemes

• Decide the proof testing intervals

• Predict the accident rate












Conclusions on Analysis Models


• Models help to visualise SIS performance

• Software speeds up analysis

• IEC 61508 part 6 - methods and tables

• Fault tree analysis for detailed systems











EIT EQO26: Unit 8 Reliability Analysis Supplementary notes on Low Demand Mode versus High Demand

Mode (also known as continuous mode)

■ Low demand mode applies when the demand on the SIS is equal to

or less than once per year. ( IEC 61511) . Alternatively no more than

two demands per proof test interval.

■ Low demand calculations use PFDavg.

■ Hazard event rate H = D x PFDavg

■ High demand mode applies when the demand on the SIS is

more than once per year. ( IEC 61511) . Alternatively more than

two

demands per proof test interval.

■ High demand mode calculations use PFH ( same as failure to danger

rate)

■ Hazard event rate H = PFH













High v Low Demand

Calculation

PFDavg = 0.05 x ½ = 0.025. and

PFH = 0.05 /8760 = 5.7E-06/hr

Suppose the demand rate D is once per year and the overpressure event rate = H/yr

In low demand mode calculation H = D x PFDavg so H = 1 x 0.025 = 0.025/yr

In high demand mode calculation H = PFH so H = 5.7E-06/hr = 0.05/yr

PSH

SIS Power

Pump Zd = 0.05 and Ti = 1/yr:


Hp safety Trip












High v Low Demand

Calculation SIS

Power

PFDavg = 0.05 x ½ = 0.025. and

PFH = 0.05 /8760 = 5.7E-06/hr

Suppose the demand rate D is once per day ( 365/yr)

And the overpressure event rate = H/yr

In low demand mode: H = D x PFDavg so H = 365 x 0.025 = 9.1/yr

In high demand mode :H = PFH so H = 5.7E-06/hr = 0.05/yr

PSH Pump

Zd = 0.05 and Ti = 1/yr:













Event rate calculation according to low or high demand mode

SIS has failures at

PFD = 0.01

PFH = 0.02/yr (2.28 E-06/hr)

Demand on SIS H = hazardous event

D = 0.1/yr ……………………………………..H = /yr ?

D = 1.0/yr ……………………………………..H = /yr ?

D = 10.0/yr ……………………………………..H = /yr ?

D = 100 /yr ……………………………………..H =


/yr ?











Documents

SIS ESD Sistems for Process Industries Using IEC 61508 Unit7 SIL Selection