Windpower Engineering & Development

  • Home
  • Articles
    • Most recent posts
    • News
    • Featured
  • Resources
    • Digital issues
    • Podcasts
    • Suppliers
    • Webinars
    • Events
  • Videos
  • 2025 Leadership
    • 2024 Winners
    • 2023 Winners
    • 2022 Winners
  • Magazine
  • Advertise
  • Subscribe

Zero Failure test, a useful statistic, can indicate system or component reliability

By Paul Dvorak | April 13, 2018

The goal of the Zero Failure test is to establish a minimum reliability for an assembly or one of its components, plus the corresponding confidence one has in this test.

John M. Berner / President and Mechanical-Industrial Engineer / Applications Research Inc.

It is important that conclusions from mechanical or electronic tests are presented unambiguously and are clearly understandable to most people.  This is particularly true for the Zero Failure test. It uses a Weibull routine, which when properly applied and clearly presented, can save much design and test time. In addition, it avoids potential misunderstandings. This same process might well be used to continuously verify the long-term performance of new multiple-unit installations. A major issue is that, at higher confidence levels, this can be a quite conservative test.

The calculation can get quite tedious. It is also easy to make mistakes if we are required to tighten reliability values.  Therefore, we’ve put together a software routine to accomplish this as part of our overall reliability program, Weibull-DR 21. Working through an example shows how to use the test and what the reveals about a product.

Suppose we have to define a required reliability for a line of 3.0 MW turbines.  How might we accomplish this?  First of all, corporate groups collectively agree to target reliability values.

For example, they might say, “We need to be 90% confident that 95% of our turbines survive the first five years under representative working or test conditions. In addition, we’ve estimated that these turbines output 6,570 megawatt-hours annually”.

This could refer to either a complete turbine or its major, most critical, or expensive components.  Over five years, this output amounts to 32,850 MWh.

Assume the analysis of previous warranty-return data indicates a Weibull Shape (or slope) factor, β = 2.5.  If you don’t have this type of history available, a good first guess for this type equipment might be between 1.5 and 2.5. Many electronic components are closer to 1.0.  Later, the Beta value could be fine-tuned from field data.

With the above, we now have enough information to solve the equation describing the target distribution. We can define the target distribution of the product or component reliability with η, the Weibull characteristic life. It is found with:

                                         η = Tmegawatt-hours / ( ln(1/R))1/ β

where Tmegawatt-hours  = target in megawatt-hours (MWh) and R = reliability value from 0 to 1.

If  Tmegawatt-hours = 32,850 (5 years), R = 0.95, and β = 2.5, then                                                                                                                                             η = 107,773 MWh, the characteristic life.

Next, we need to satisfy the confidence requirement. In this case, Confidence, C = 90% (to minimize the possibility of a false-positive test).

 

 

 

 

where n = sample size, f = failures allowed, and Rtest = the reliability corresponding to the required test in MWh. Its value is determined by iteration to satisfy the above equation.

For our sample, n = 3, f = 0, Rtest = (1 – C)1/n, and   Confidence, C = 0.90, the reliability of the test.

First, calculate Rtest using the figures above, β = 2.5, and η = 107,773:

                              Rtest = 0.4642.

Now we back-calculate to get the test’s Weibull parameters.   

       Tmegawatt-hours  = η * (Ln(1/Rtest)1/β                                                                                                                                                                                                  = 96,957 MWhours,

So the zero failure test reveals that a turbine will produce 96,951 MWh-per-unit without failures.

The table lists results from the zero-failue test.

Assuming 25% capacity factor, we can run 3.0 MW, 90% of full time to accelerate the test.  Using three test cells, we should be able to complete this test, exclusive of the blades, within a few months.

Summary
You can be 90% confident that a sample passing this test represents an overall population, the reliability of which exceeds 95% at an elapsed period of 32,850 MWh, or approximately five years.

The graphic above, a report, was generated by the software routine for the values presented here.

For the latest Weibull-DR: www.weibulldr.com or

jmb@applicationsresearch.com

 

 


Filed Under: Projects
Tagged With: applications research inc
 

About The Author

Paul Dvorak

Related Articles Read More >

US government allows Empire Wind offshore project to resume construction
Overlooked and underleveraged: Why ‘lite repowering’ is wind energy’s best near-term bet
79 aging wind turbines brought back online throughout Texas panhandle
Data center signs 166-MW PPA with Las Majadas Wind in Texas

Podcasts

Wind Spotlight: Looking back at a year of Thrive with ZF Wind Power
See More >

Windpower Engineering & Development Digital Edition

Digital Edition

Browse the most current issue of Windpower Engineering & Development and back issues in an easy to use high quality format. Clip, share and download with the leading wind power engineering magazine today.

Windpower Engineering & Development
  • Wind Articles
  • Solar Power World
  • Subscribe to Windpower Engineering
  • About Us/Contact Us

Copyright © 2025 WTWH Media LLC. All Rights Reserved. The material on this site may not be reproduced, distributed, transmitted, cached or otherwise used, except with the prior written permission of WTWH Media
Privacy Policy | Advertising

Search Windpower Engineering & Development

  • Home
  • Articles
    • Most recent posts
    • News
    • Featured
  • Resources
    • Digital issues
    • Podcasts
    • Suppliers
    • Webinars
    • Events
  • Videos
  • 2025 Leadership
    • 2024 Winners
    • 2023 Winners
    • 2022 Winners
  • Magazine
  • Advertise
  • Subscribe