Reliability HotWire: eMagazine for the Reliability Professional
Reliability HotWire

Issue 20, October 2002

Hot Topics

Determining the Reliability of a System with Load Sharing

A reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. Meaning, the failure of component A does not effect the failure of component B. However, if your system consists of multiple components sharing a load then the assumption of independence no longer holds true. If one component fails then the component(s) that are still operating will have to assume the failed unit's load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. This article will explore the concept of component dependence while using accelerated life test data to determine system reliability.

In the case of load-sharing components, it is required to know the change of the failure distributions of the surviving components in order to determine the system's reliability. Accelerated test data and analysis can be used to determine the failure distribution of each component at different stress levels.

Example

Consider a system of two units connected reliability-wise in parallel, which must supply an output voltage of 8 volts. If both units are operational, then each component is to generate 50% of the total output. What is the system reliability at 8760 hours (1 year)? The system is presented in Figure 1.

System of two units connected reliability-wise in parallel

Figure 1: System of two units connected reliability-wise in parallel

However, if for example Unit 1 fails, then Unit 2 will assume the responsibility of supplying the entire load by itself. This obviously will have a great impact on the reliability of Unit 2 since it is now being asked to supply twice the output. This will continue for as long as Unit 1 is down. Note: For this example, we will only consider the non-repairable case, i.e. when a component fails, it is not repaired/replaced. While this is occurring, the probability of Unit 2 failing due to the increased load (stress) has also increased (compared to when the two units share the load equally).

The first step is to determine the failure distribution for each unit. This can be done using accelerated test data. A total of 20 units were tested to failure at 7, 10, and 15 volts. The accelerated test data is presented in Table 1.

Table 1: Accelerated test data

Time-to-Failure

(hrs)

Voltage

(volts)

874 7
2253 7
3026 7
3115 7
3575 7
3918 7
5000 7
5290 7
340 10
551 10
560 10
825 10
1079 10
1140 10
1701 10
2800 10
105 15
167 15
275 15
362 15

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components may not be the same. If that were the case, data would need to be collected for each component.

The data in Table 1 was analyzed using ReliaSoft's ALTA 6 with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution. The estimated parameters, β, K, and n, are shown next.

Beta = 1.9239, K = 3.2387E-7, n = 3.4226

Now that the failure properties have been determined using the accelerated test data, the reliability of the system at some time, t, can be calculated using the following equation.

Equation 1 - R(t) = Rsub1(t, Ssub1) x Rsub2(t, Ssub2)

(1)

where:

Ssub1 = Psub1 x S and Ssub2 = Psub2 x S

P1 and P2 represent the portion of the total load that each unit is to supply when both units are operational, S1 and S2 represent the load share that Unit 1 and Unit 2 must support, respectively, when both units are operational, and S is the total required output.

t1e is the equivalent operating time for Unit 1 if it had been operating at S instead of S1. A graphical representation of te is shown in Figure 2, where the red line represents the low stress (load) and the green plot line represents the high stress (load).

Graphical representation of te in a Reliability vs. Time plot

Figure 2: Graphical representation of te

t1e can be calculated by:

Equation - Rsub1(t) = Rsub1(tsub1e)

t2e is calculated in the same manner:

Equation - Rsub2(t) = Rsub2(tsub2e)

In this example, the reliability equations for Unit 1 and Unit 2 are equivalent since they are the same type of component and demonstrate the same failure properties. In addition, the total output is split equally between the two units (when both units are operating), therefore t1e and t2e will also be the same. So once all of the pieces are in place, then the system reliability, R(t=8760), can be calculated using Eqn. (1). The resulting system reliability is presented next.

R(8760) = 0.8567

Therefore, the reliability of the system, assuming a total output of 8 volts is 85.67%. Fortunately, you will not be required to conduct these calculations by hand. The upcoming release of BlockSim 6 will include the ability to calculate system reliability while considering load sharing. Additional information on the BlockSim can be found at http://BlockSim.ReliaSoft.com/.

 

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