PS2 PS1
Availability
Availability
For successful operation on demand, the availability, A is
given by
A = MTBF / (MTBF + MDT)
where
MTBF= ' Mean Time Before Failure' = 1/FR
MDT = Mean Dead Time (assumed to be equal to 0.5 x
Proof Test Interval, T)
FR = Unrevealed or fail-safe failure rate.
AT = ?
Equal Availability
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For systems with equal availability
For n =2:
AT
For 2002, Total Availability:
For n = 3:
For 1002, Total Availability:
Example 1: Equal Availability
PS PS
PS
PS
1002 ESD
ESDV Fig. 2: Availability Diagram
Fig. 1: Equal Availability System
Device FR x 10
6/hr Proof Test Interval
(months)
PS 3 6
ESDV 6.1 6
and
and,
Therefore for the whole system the total availability is:
Systems of Unequal Availability
a b Availability
Y Y A1A2
Y N A1(1-A2)
N Y (1-A1)A2
b
Availability Diagram for Systems of Unequal Availability
A1
a
A2
x
Example 2: Systems of Unequal Availability
Process flow diagram
PS1
(A1)
PS2
(A2)
½
or
2/2
ESDV
(A3)
Availability block diagram
PS2 PS1
Systems of Unequal Availability



MDTMTBF
MTBF
A
F
1
IntervalTest Proof5.0
F
1
R
R

10
365
30
5.0
10
1

A
5
365
20
5.0
5
2

A
5.2
365
60
5.0
5.2
3

A
Given that,
96817.03 A99455.02 A99591.01 A
Device FR /yr Proof Test
Interval
(days)
PS1 0.1 30
PS2 0.2 20
ESDV 0.4 60
Since systems of unequal availability, use truth table:
a b
Y Y A1A2
Y N A1(1-A2)
N Y (1-A1)A2
Now for 1002 voting system;
1002 voting => APS = A1A2+A1(1-A2)+(1-A1)A2
APS= 0.99591 x 0.99455 + 0.99591 x (1-0.99455)
+ (1-0.99591) x 0.99455
= 0.99998
For 2002 voting system =>
APS = A1A2 = 0.99591 x 0.99455 = 0.99048
Hence, Atotal = APSA3 = 0.99048 x 0.96817
= 0.95895 (for 2002)
Atotal= 0.99998 x 0.96817= 0.96815 (for 1002)
Note: Once again the 1002 voting system provides a higher degree of
protection but at the cost of increased likelihood of spurious signals.
Also it is noteworthy that the system with equal availability pressure switches
provides a higher overall availability as compared to the one with unequal
availabilities.
Example 3: System of Equal Availability
A storage vessel is served by 3 pressure safety relief valves (PSV) all
set at the same pressure. In the event of the system exceeding the
system set pressure, for a 2003 voting PSV assembly, calculate the
overall trip availability in successfully isolating the flow to the storage
vessel using an emergency shutdown valve (ESDV). You may use the
following data:
PSV ESDV
Failure rate (/yr) 10-2 10-3
Proof Test
Interval (yr)
5 10
The Mean Dead Time is assumed to be equal to 0.5 x Proof Test Interval
MTBF
(Yr)
MDT
(Yr)
A
PS 102 2.5 0.97561
ESDV 103 5 0.99502
Schematic Representation of the System
MDTMTBF
MTBF
A


MTBF= ' Mean Time Before Failure' = 1/FR
MDT = Mean Dead Time (assumed to be equal to 0.5 x
Proof Test Interval, T)
FR = Unrevealed or fail-safe failure rate.
Storage Vessel
ESDV
PS PS PS
PS
PS
PS
ESDV
A=0.97561
A=0.97561
A=0.97561
A=0.99502
2003
Availability Block Diagram
The Availability for the PSV
system is:
ESDVTPSVSYSTEM
AAA 
Example 4:
In order to minimise the risk of a highly exothermic runaway
reaction occuring in a gas phase reactor, the system is
equipped with three Temperature Switches (TS) set to activate
an Emergency Shutdown Valve (ESDV) to terminate the feed
flow into the reactor in the event of a rapid rise in the
temperature.
To prevent an accompanying rise in pressure, a Pressure
Switch (PS) opens a Pressure Relief Valve (PRV) to reduce the
pressure.
For the above system:
a) draw the corresponding a simple Process & Instrumentation
flow Diagram (P&ID)
b) draw the corresponding individual and overall availability
block diagrams
c) calculate the overall system availability for a 2003 voting
system for the pressure switches to protect against both a rise
in temperature and pressure using the data given in Table1
Component
Failure rate
(yr-1)
Proof
test
interval
(yr)
Temperature Switch (TS) 2.5 x10-2 1
Pressure Switch (PS) 2.0 x 10-2 1
Emergency Shutdown
Valve (ESDV)
5.5 x 10-3 6
Pressure Relief Valve
(PRV)
5.5 x 10-3 6
Table 1: Failure rate and proof test interval data
APS
APRV
Emergency Shutdown Valve (ESDV):
Pressure Relief Value / Pressure Switch:
The overall availability block diagram for the entire reactor is
then,
For calculation of the availability for pressure and temperature
switches and the valves:
A = MTBF / (MTBF + MDT) (1)
MTBF = ‘Mean Time Before Failure’ = 1/Failure rate (2)
MDT = ‘Mean Dead Time’= Proof test interval/2 (3)
ATS = (1/ 2.5 x 10
-2)/( (1/ 2.5 x 10-2)+1/2) = 0.9877 (4)
APS = (1/2.0 x 10
-2)/( (1/ 2.0 x 10-2)+1/2) = 0.9901 (5)
AESDV = (1/ 5.5 x 10
-3)/( (1/ 5.5 x 10-3)+6/2) = 0.9838 (6)
APRV = (1/ 5.5 x 10
-3)/( (1/ 5.5 x 10-3)+6/2) = 0.9838 (7)
Failure
rate (yr-1)
Proof test
interval (yr)
Availability
TS 2.5 x 10-2 1 0.9877
PS 2.0 x 10-2
1 0.9901
ESDV 5.5 x 10-3
6 0.9838
PRV 5.5 x 10-3
6 0.9838
In summary:
For the PRV system consisting of the pressure switch and the PRV,
Asystem_PRV = APSAPRV=0.97410 (8)
For the ESDV system consisting of the temperature switch (2003
voting) and the ESDV,
Athree_TS = ATS
3+3(1-ATS)ATS
2 = 0.9995 (9)
Asystem_ESDV = Athree_TSAESDV = 0.9834 (10)
The overall availability of the entire reactor is then,
Asystem_overall = Asystem_PRV x Asystem_ESDV = 0.9578
AY1 0.98700
0.99997 0.98700
Section (i) Section (ii)
0.99356
LS
LS
HIPS
1/2
SOV 2
EY20012B
ESD 2
SOV 1
EY20012A
1/2
ESDV
20012
0.99250 0.99997
0.99250
Availability Diagram
Example 5: Calculate the total Availability for the
following system:
Therefore, Total Availability for Section (i) = 0.99994
Availability Diagram for Section (i)
LS/
HIPS
LS
A1=0.99247
A2=0.99250
The Availability for Part Y1 is:
a b
Y Y = A1 A2 = 0.99247 x 0.99250 = 0.98503
Y N = A1 (1-A2) = 0.99247 x (1-0.99250) = 7.44 x 10
-3
N Y = (1-A1) A2 = (1-0.99247) x (0.99250) = 7.47 x 10
-3
Truth Table for Section (i)
Ay1 = 0.99250 x 0.99997 = 0.99247
SOV 2
ESD/
SOV1
A1=0.98700
A2=0.98697
For Section (ii):
a b
Y Y = A1 A2 = 0.98700 x 0.98697 = 0.97414
Y N = A1 (1-A2) = 0.98700 x (1-0.98697) = 0.01286
N Y = (1-A1) A2 = (1-0.98700) x 0.98697 = 0.01283
Availability Diagram for Section (ii)
Truth Table for Section (ii)
Therefore, the total availability for Section (ii) = 0.99983
Therefore, the total availability,
AT = 0.99983 x 0.99994 x 0.99356
= 0.99331