Standby System
In previous numerical examples and in most industrial applications (cases studies), all the components within the system are supposed to be independent. For ex- ample, the failure of component A does not affect the failure of component B.
Consider two pumps, Pump1 and Pump2, in a standby redundancy system. For each block of the system the “active” failure distribution is distin- guished by the “quiescent” failure distribution. In particular, the quiescent failure distribution refers to the component when it is in standby mode.
For a generic component the failure modes during the quiescent mode are generally different from those during the active mode.
In the case of identical failure distributions for both quiescent and active modes, the components are in a simple parallel configuration (also called a “hot standby” configuration). When the rate of failure of the standby component is less in quiescent mode than in active mode, then the configuration is called a “warm standby” configuration. Lastly, in a cold standby con- figuration the rate of failure of the standby component is zero in quiescent mode (i. e., the component cannot fail when in standby).
Dealing with standby systems, a switching device to the standby component in the case of failure for the active component is often present. In particular, it is possible for the switch to fail before the active compo- nent. If the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
6.9.1.1 Nonrepairable Components, Exponential Distribution of ttf. Perfect Switch
Figure 6.66 presents the trend of F .t /, R.t /, f .t /, and.t /for different values oft, distinguishing and comparing the hot standby system (where both quies- cent and active failure distributions are the same – first column in the figure) from the cold standby system (where the rate of failure of the standby component is zero in quiescent mode – second column in the figure).
Both systems are supposed to be not repairable. Obvi- ously, as demonstrated by Fig. 6.66, the cold system
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Block Unreliability vs Time
Time, (t)
Unreliability, F(t)=1-R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 1.000
0.200 0.400 0.600 0.800
Unreliability Diagram1
System Stand-by 1 (Pump2 in stand by)
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Block Unreliability vs Time
Time, (t)
Unreliability, F(t)=1-R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 1.000
0.200 0.400 0.600 0.800
Unreliability Diagram1
System Stand-by 1 (Pump2 in stand by)
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Block Reliability vs Time
Time, (t)
Reliability, R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 1.000
0.200 0.400 0.600 0.800
Reliability Diagram1 System Stand-by 1 (Pump2 in stand by)
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Block Reliability vs Time
Time, (t)
Reliability, R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 1.000
0.200 0.400 0.600 0.800
Reliability Diagram1 System Stand-by 1 (Pump2 in stand by)
ReliaSoft BlockSim 7 - www.ReliaSoft.com
Block Probability Density Function
Time, (t)
f(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000 0.000
7.000E-5
1.400E-5 2.800E-5 4.200E-5 5.600E-5
Pdf Diagram1
System Stand-by 1 (Pump2 in stand by)
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Block Probability Density Function
Time, (t)
f(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000 0.000
5.000E-5
1.000E-5 2.000E-5 3.000E-5 4.000E-5
Pdf Diagram1
System Stand-by 1 (Pump2 in stand by)
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Block Failure Rate vs Time
Time, (t)
Failure Rate, f(t)/R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000 0.000
2.000E-4
4.000E-5 8.000E-5 1.200E-4 1.600E-4
Failure Rate Diagram1
System Stand-by 1 (Pump2 in stand by)
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Block Failure Rate vs Time
Time, (t)
Failure Rate, f(t)/R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000 0.000
1.000E-4
2.000E-5 4.000E-5 6.000E-5 8.000E-5
Failure Rate Diagram1
System Stand-by 1 (Pump2 in stand by)
Fig. 6.66 Hot standby versus cold standby. Pump1active, Pump2 standby. Exponential distribution. Nonrepairable components:
F .t /,R.t /,f .t /,.t /. ReliaSoft®software
is better than the hot one because the standby com- ponent is “as good as new” till the switch component, supposed to be perfect, switches the active and the qui- escent pumps (i. e., it substitutes the originally active component which fails).
6.9.1.2 Nonrepairable Components, Mix of Probability Distributions of Blocks’ ttf. Not Perfect Switch
Similarly to the analysis conducted in the previous sec- tion, Fig. 6.67 presents the trend ofF,.t /andf .t / for different values oft, distinguishing and comparing the hot standby system (where both quiescent and ac- tive failure distributions are the same – first column of figure) from the cold standby system (where the rate of failure of the standby component is zero in quiescent mode – second column of figure), and assuming:
Block Unreliability vs Time
Time, (t)
0,000 10000,000 20000,000 30000,000 40000,000 50000,000
0,000 1,000
0,200 0,400 0,600 0,800
Unreliability Diagram1
System Stand-by 1 (Pump2 in stand by)
Block Reliability vs Time - Cold stand by
Time, (t)
0,000 10000,000 20000,000 30000,000 40000,000 50000,000
0,000 1,000
0,200 0,400 0,600 0,800
Reliability Diagram1 System Stand-by 1 (Pump2 in stand by)
Block Reliability vs Time
Time, (t)
0,000 10000,000 20000,000 30000,000 40000,000 50000,000
0,000 1,000
0,200 0,400 0,600 0,800
Reliability Diagram1 System Stand-by 1 (Pump2 in stand by) ReliaSoft BlockSim 7 - www.ReliaSoft.com
Block Failure Rate vs Time - Cold stand by
Time, (t)
Failure Rate, f(t)/R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
-2.000E-5 3.000E-4
4.400E-5 1.080E-4 1.720E-4 2.360E-4
Failure Rate Diagram1
System Stand-by 1 (Pump2 in stand by)
ReliaSoft BlockSim 7 - www.ReliaSoft.com
Block Probability Density Function
Time, (t)
f(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 8.000E-4
1.600E-4 3.200E-4 4.800E-4 6.400E-4
Pdf Diagram1
System Stand-by 1 (Pump2 in stand by)
ReliaSoft BlockSim 7 - www.ReliaSoft.com
Block Probability Density Function - Cold stand by
Time, (t)
f(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
-3.000E-6 2.000E-4
3.760E-5 7.820E-5 1.188E-4 1.594E-4
Pdf Diagram1
System Stand-by 1 (Pump2 in stand by) Block Unreliability vs Time Cold stand by
Time, (t)
Unreliability, F(t)=1-R(t)
0,000 10000,000 20000,000 30000,000 40000,000 50000,000
0,000 1,000
0,200 0,400 0,600 0,800
Unreliability Diagram1
System Stand-by 1 (Pump2 in stand by) ReliaSoft BlockSim 7 - www.ReliaSoft.com
Block Failure Rate vs Time
Time, (t)
Failure Rate, f(t)/R(t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
0.000 0.002
4.000E-4 8.000E-4 0.001 0.002
Failure Rate Diagram1
System Stand-by 1 (Pump2 in stand by)
Fig. 6.67 Hot standby versus cold standby. Pump1active, Pump2standby. Nonrepairable components:F .t /,R.t /,f .t /,.t /.
Switch not perfect. ReliaSoft®software
• Pump1. Exponential distribution, MTTFPump1 D 10;000 h;
• Pump2. Normal distribution, MTTFPump2D6;000 h, and standard deviation of ttf 100 h;
• Switch. Weibull distribution, scale parameter˛ D 7;000 h, and shape parameterˇD1:5.
Both hot and cold time-dependent systems are sup- posed to be not repairable.
6.9.1.3 Nonrepairable Components and Simulation Analysis.
Hot Standby System and Switch Perfect The following analysis was conducted with the use of Monte Carlo simulation in order to test the system be- havior in accordance with the hot and cold hypotheses.
In particular, Fig. 6.68 presents an up/down diagram
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Block Up/Down
Time, (t)
0.000 10000.000 20000.000 30000.000 40000.000 50000.000
System Pump 1 Pump 2 Stand-by 1 (Pump2 in stand by) Stand-by 1 (Pump2 in stand by)_Switch
Fig. 6.68 Hot standby, simulation analysis. Switch perfect. ReliaSoft®software related to the nonrepairable hot standby system made
of pumps Pump1 and Pump2, and a “perfect” switch component, i. e., a component which does not fail and it is not subject to failures. From Fig. 6.68, the standby system fails when the active Pump1fails because non- repairable Pump2 fails first, i. e., during the standby period.
6.9.1.4 Nonrepairable Components and Simulation Analysis, Cold standby system
Figure 6.69 presents the up/down diagram obtained by a simulation analysis. It shows the system failing when Pump1fails because the switch fails first, i. e., it is in the state of failure when Pump1fails and has to be sub- stituted by Pump2.
If the switch is perfect, Pump2action starts imme- diately when Pump1fails as illustrated in Fig. 6.70.
6.9.1.5 Repairable Components and Simulation Analysis. Hot Standby System
and Switch Perfect
Assuming an exponential distribution of ttr (MTTRPump1 DMTTRPump2 D100 h), a Monte Carlo simulation analysis generates the state diagram shown
in Fig. 6.71 for the hot standby system. Figure 6.72 reports the trend of the expected availability and reliability of the hot standby system as the result of a simulation analysis by ReliaSoft® reliability software.
Figure 6.73 shows the results of the simulation analysis with MTTR equal to 1,000, it can be stated that the system too passes from up to the down when it fails because both Pump1 and Pump2are under the random repair process (between 20,000 and 30;000 h).