An article in IEEE Transactions on Dielectrics and

Chapter 4, Problem 169E

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QUESTION:

Problem 169E

An article in IEEE Transactions on Dielectrics and Electrical Insulation [“Statistical Analysis of the AC Breakdown Voltages of Ester Based Transformer Oils” (2008, Vol. 15(4))] used Weibull distributions to model the breakdown voltage of insulators. The breakdown voltage is the minimum voltage at which the insulator conducts. For 1 mm of natural ester, the 1% probability of breakdown voltage is approximately 26 kV, and the 7% probability is approximately 31.6 kV. Determine the parameters δ and β◊of the Weibull distribution.

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QUESTION:

Problem 169E

An article in IEEE Transactions on Dielectrics and Electrical Insulation [“Statistical Analysis of the AC Breakdown Voltages of Ester Based Transformer Oils” (2008, Vol. 15(4))] used Weibull distributions to model the breakdown voltage of insulators. The breakdown voltage is the minimum voltage at which the insulator conducts. For 1 mm of natural ester, the 1% probability of breakdown voltage is approximately 26 kV, and the 7% probability is approximately 31.6 kV. Determine the parameters δ and β◊of the Weibull distribution.

ANSWER:

Solution :

Step 1 of 1:

Let X denotes break voltage with weibull distribution.

Given the 0.01 probability of breakdown voltage is approximately 26 kV, and the 0.07 probability is approximately 31.6 kV.

Our goal is:

We need to determine the parameters δ and β of the Weibull distribution.

Now we need to determine the parameters δ and β of the Weibull distribution.

P(X26) = 0.01 and P(X31.6) = 0.07.

P(Xx) = 1-

First we are finding P(X26) = 0.01.

P(X26) = 1-= 0.01

P(X26) = = 1-0.01

P(X26) =

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