The mechanical effects of short-circuit currents in - Montefiore

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Only the lowest break values are important for determination of the reliability. It is interpreted as a quasi-incomplete sample test [Ref 98]. As a result it is the minimum breaking load of the brand-new post insulator with a minimum breaking load of 8 kN on the 0,2-%-quantil (Figure 5.4, point A). For the 25year-old post insulators the minimum breaking load (6 kN) is on the 3-%-quantil (Figure 5.4, point B). 5.2.1.4 Reliability of 110 kV post insulators depending on mechanical stress and strength (example) 5.2.1.4.1. Design practice The mechanical reliability of a 110 kV post insulator results from the comparison of stress and strength. If the stress is greater than the strength the insulator will be damaged. For substation design the highest possible shortcircuit current at the end of the substation life-time has to be considered. This loading multiplied by a safety factor (i.e. [Ref 56] recommends a maximum load not higher than 70 % of the minimum breaking load) is the criterion for selection of the insulators. This means that the manufacturer-guaranteed strength F umin is higher than the maximum stress which will be achieved in the future. For design, both reliability influencing factors (stress and strength) are considered as one point distributed. It is assumed that the highest possible current appears in each shortcircuit and that each post insulator cannot be damaged by loads lower than the minimum breaking load. This procedure is well tested in practice. If the mechanical reliability of a device must be determined exactly (i.e. for short-circuit current level uprating in a substation) the method of consideration must be changed. 5.2.1.4.2. Reliability calculation for a 110 kV post insulator The reliability Z of a device results from the failure probability p f : (5.5) Z = 1 − pf The probability of a mechanical failure depends on the stress density function p B and on the strength distribution function V S. If the stress density function and the strength distribution function are known the failure probability can be calculated by solving the Duhamel integral : 78 Fmax p = ∫ p F . V F . dF (5.6) ( ) ( ) f B S Fmin For the given example (disconnector equipped with 110 kV post insulator) the stress density function results from the load probability and the mechanicaldynamical behaviour of the system "conductorsupport» due to short-circuit current loading [Ref 57]. For this example the relationship determined by experiments is known between – short-circuit current loading and stress-maximum – static load and stress Therefore it is possible to convert the load due to a certain short-circuit current loading to a static load at the top of the post insulator which produces the same stress. The stress density function results then (with the conversion) from the calculated density function of the short-circuit current. For the given example the strength distribution function was derived from the part-collectives with the lowest break loads. Fault probabilities of p f = 5,3.10 -13 for the brand-new post insulator with a minimum breaking load of 8 kN and of p f = 4.10 -9 for the 25-year-old post-insulators with a minimum breaking load of 6 kN result by solving the integral (5.6) for the example. 5.2.1.4.3. Stress and strength influences on probability Provided that the course of curves of stress and strength are constant the probability depends on stress and strength as follows : 1. If the stress Fmax increases about 10 % of the value of the minimum breaking load the fault probability will rise tenfold. 2. If the strength Fumin of a post insulator is 10 % lower than the minimum breaking load (i.e. Fumin = 7,2 kN instead 8 kN) under the same load, the fault probability will also rise tenfold. That means that the same percentage increase of stress or decrease of strength leads to the same decrease of probability. For the given example (disconnector equipped with 110 kV post insulators) there is good co-ordination between stress and strength concerning the influence of fault probability: the gradient of fault prohability is determined equally by both because the stress-increasing gradients and the strength-decreasing gradients are the same. For practical cases it can be expected that the loading and the stress increase during the life-time of a device. But the strength normally decreases. In such a
case, the reduction of reliability results from the combination of stress-increasing and strengthdecreasing effects. In the example, the stress of a post insulator with a minimum breaking load of 8 kN increased from 5,6 kN to 6,4 kN. At the same time the strength decreased from 8 kN to 7,2 kN. As a result of increasing stress and decreasing strength the failure probability rose by two decades. If the relative changes of stress and strength are unequal, the probability will be taken as follows : – For a given permissible failure probability one can find parameter combinations Fmax/Fmin which do not exceed the given probability limit. – If an equivalent static load Fmax is assumed, it will be possible to determine minimum strength of post insulators with which an assumed limit of failure probability will be observed. 5.2.1.5 Summary In the described example of mechanical reliability of 110 kV post insulators, it was shown that if the stress density function and the strength distribution function are known, the reliability can be calculated. If the percentage increase of stress and the percentage decrease of strength as the same, the influences on the reliability will be equal. A good co-ordination of effects of increasing stress and decreasing strength of the reliability results in the example considered. For other devices as well, it should be assumed that the reliability is influenced by increasing stress and decreasing strength. Figure 5.5 Load / strength distributions for rigid bus example 79 Figure 5.6 Bending breaking load Figure 5.7 Bending breaking load 5.2.1.6 Example of savings using probabilistic approach in renovation (Ford, Sahazizian) Type of shortcircuit Nominal voltage 110 kV 220 kV 400 kV 500 kV 1-ph + e 70 % 85 % 90 % 92 % 2-ph 10 % 6 % 3 % 2 % 2-ph + e 15 % 5 % 4 % 3 % 3-ph 5 % 4 % 3 % 3 % Σ 100 % 100 % 100 % 100 % Table 5.1 Various types of short-circuits (contributions) Courtesy M. Daszczyszak, University of Mining and Metallurgy, Cracow, Poland.
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THE MECHANICAL EFFECTS OF SHORT-CIR
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3.7. Special problems .............
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1. INTRODUCTION 1.1. GENERAL PRESEN
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The pinch (first maximum) can be ve
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1.3.3.2 Supporting structures Simil
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This method allows to state the sho
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and becomes the static force in equ
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(2.26) M el-pl ⎡ y d/ 2 y = 2⎢
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V σ 3 2,5 2 1,5 1 mechanical reson
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V F 3 2,5 2 1,5 1 mechanical resona
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1. eigenmode 2. eigenmode 3. eigenm
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(2.43) U E max max = 1 2 1 = 2 l
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m (2.47) M = σm = Z mσ m dm 2 J w
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Page 35 and 36: d)Methods used IEC 60865 method : -
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Page 39 and 40: In Figure 3.5 to Figure 3.8, the ma
Page 41 and 42: Figure 3.17 Comparison calculated a
Page 43 and 44: movement of the dropper (swing out
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Page 57 and 58: Force / kN Force / kN 100mm 200mm 4
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Page 61 and 62: 3.7. SPECIAL PROBLEMS 3.7.1. Auto-r
Page 63 and 64: for the maximum outswing and the ma
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Page 104 and 105: Figure 6.11 Coefficients mJs1 and m
Page 106 and 107: 7. REFERENCES Ref 1. IEC TC 73/CIGR
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Page 110 and 111: Ref 80. Fiabilité des structures d
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Page 118 and 119: Figure 8.9 Case 3 (third eigenfrequ
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8.4. INFLUENCE OF THE RECLOSURE The
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( ( δ ) ( δ ) ) T = Mg08 . b r .
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IEC 60865 Calculations If the clear
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8.6.1.1 With calculation of the rel
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f) Calculation of the stresses due
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d) Determination of the forces at t
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8.7. ERRATA TO BROCHURE NO 105 In V
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