Can frequent rotating equipment switchovers reduce the life of the motors and mechanical parts?

Hi josh,
Good question !

I think yes. I havebeen experienec in my refinery since 12 years that frequent changing over of standby machine reduces the life of the seal, bearing and other components. As per record statistics increasing the switch over schedule from monthly to 6 month my failure rates have been reduced by 30 %

waht will be th eright to to fix switch over schedule , can any one advice.

regards

There are a few ways of looking at it. We have been involved in strategies related to alternating redundant systems for some time.

One of the topics that comes up is that the equipment will somehow fail at the same time if they are operated equally. That is actually not the case as most failures are 'random' - meaning that there are certain tolerances in each machine and while they may have a similar mean life time, they are not identical. As the life of the machinse continue, their actual total run time will differ based upon when they were last overhauled/repaired.

The real questions should be:

1. How long does the combined system last?
2. Are there idle failures that exist (ie: false brinelling of bearings, seal failure, winding insulation contamination/moisture adsorption, etc.) such that the total run time of the spare is far less than that of the primary (primary runs all the time, spare is idle scenario)? Or,
3. Can the complete system reliability be improved through an alternating strategy that is not 50:50?

There really is not a generic direct answer to these questions. It is actually dependant upon the individual system - which is one of the lessons that the original work into Reliability-Centered Maintenance was all about because, even in a redundant system there are slight differences between each machine.

I include an example of a redundant system, that is based upon a real system we looked at a few years ago, in a white paper entitled "Considerations for Time To Failure Estimation" which I have attached for comment.

Sincerely,
Howard

Considerations_in_Time_to_Failure_Estimation.pdf (125 Kb, 23 downloads)

Motodoc,

I try to understand your paper by doing the calculation myself as follows:

For seal, MTTF= (6000x5 - (96x2))/2 = 14904
so failure rate f = 1/MTTF = 1/14904 = 6.7 x 10E-6.

For bearing failure of the primary motor, f = 1/22000 = 4.5 x 10E-5.

For bearing failure of spare motor, f = 1/ 13500 = 7.4 x 10E-5.

For winding failure of spare motor, f = 1/20000 = 5 x 10E-5.

How did you get operating failure rate of 6.9 x 10E-3? I got MTTF = (6000x5 - (96x3))/3 = 9904 so f = 1/9904 = 1.0 x10E-4.

This message has been edited. Last edited by: Josh,

On the operating failure rate:

In this case, the failure rate is based upon the actual operating time of the second pump, or 2 failures over 288 hours = 6.9 x 10^-3. I did not include the idle time in the calculation to demonstrate the impact of the actual operating hour difference between the two machines.

I suppose the main point of that part of the paper is that the total reliability of the redundant system should be considered and a strategy based upon that system should be realized.

Thanks for your comments!

Howard

Ok, then it is then MTTF = 288/2 = 144 so f=1/144 = 6.9x 1oE^3.

I will try to get your graph.