BECKHOFF TwinSAFE - Manual (Page 130)

Circuit examples
128 Application Guide TwinSAFE
Inserting the values, this produces:
S1:
݊
௢௣
ൌ
230 כ 8 כ 60
60
ൌ 1840
ܯܶܶܨ
ௗ
ൌ
1000000
0.1 כ 1840 ൌ 5434.8y ൌ 47608848h
S2:
݊
௢௣
ൌ
230 כ 8 כ 60
60
ൌ 1840
ܯܶܶܨ
ௗ
ൌ
2000000
0.1 כ 1840 ൌ 10869.6y ൌ 95217696h
and the assumption that S1 and S2 are in each case single-channel:
ܯܶܶܨ
ௗ
ൌ
1
ߣ
ௗ
results in a
ܲܨܪ ൌ
0.1 כ ݊
௢௣
כ ሺ1 െ ܦܥሻ
ܤ10
ௗ
ൌ
1 െ DC
MTTF
ୢ
S1:
ܲܨܪ ൌ
1 െ 0.99
5434.8 כ 8760
ൌ 2.10E െ 10
S2:
ܲܨܪ ൌ
1 െ 0.99
10869.6 כ 8760
ൌ 1.05E െ 10
The following assumptions have to be made now:
The door switches S1/S2 are always actuated in opposite directions. Since the switches have different
values, but the complete protective door switch consists of a combination of break and make contacts
and both switches must function, the poorer of the two values (S1) can be taken for the combination!
There is a coupling coefficient between the components that are connected via two channels. Examples
are temperature, EMC, voltage peaks or signals between these components. This is assumed to be the
worst-case estimation, where ß =10%. EN 62061 contains a table with which this ß-factor can be
precisely determined. Further, it is assumed that all usual measures have been taken to prevent both
channels failing unsafely at the same time due to an error (e.g. overcurrent through relay contacts, over
temperature in the control cabinet).
This produces for the calculation of the PFH value for block 1:
PFH
tot
= β* (PFH(S1)+ PFH(S2))/2 + PFH(EL1904) + PFH(EL6900) + PFH(AX5805) +
PFH(AX5805) + PFH(AX5805) + PFH(AX5805)
to:
PFH PFH PFH PFH
tot tot tot tot
= == =
10%*
(2.10E-10+1.05E-10) / 2 +1.11E-09 + 1.03E-09 + 5.15E-09 + 5.15E-09 +
5.15E-09 + 5.15E-09 = 2.28E 2.28E 2.28E 2.28E----08 08 08 08