The mechanical effects of short-circuit currents in - Montefiore

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Symmetrical associated-phase layout bI bII phase 1 phase 2 phase 3 phase 1 phase 2 phase 3 i1 i2 i3 i1 i2 i3 Figure 2.25 b=busbar The following expression of Laplaces’s force on phase 2 of busbar 1 corresponds to the symmetrical associated-phase layout represented above : µ 0 ⎛ i1 i3 i3 i2 i1 ⎞ (2.90) F = . i2. ⎜ − + + + ⎟ 2π ⎝ d d D + d D + 2d D + 3d ⎠ Type of fault phase to phase fault developing through one busbar two-phase fault developing through two busbars phase to earth fault developing through two busbars three-phase fault developing through two busbars block diagram analyzed b 1 p 1 I b 1 p 2 I b 1 p 3 b 2 p 3 b 2 p 2 b 2 p 1 b 1 p 1 I b 1 p 2 I b 1 p 3 b 2 p 3 I b 2 p 2 I b 2 p 1 b 1 p 1 b 1 p 2 b 1 p 3 I b 2 p 3 I b 2 p 2 b 2 p 1 b 1 p 1 I1 b 1 p 2 I2 b 1 p 3 I3 b 2 p 3 I3 b 2 p 2 I2 b 2 p 1 I1 d d D d d 34 1,4 1,2 1 0,8 0,6 0,4 0,2 SYMMETRICAL ASSOCIATED PHASES amplification coefficients 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 RATIO d/D K 2p 1 b K 3p 1b K 1p 2 b K 2p 2 b K 3p 2 b Figure 2.26 Comparison of stresses versus type of faults Amplification Coefficients Maximum Asymmetry 1b K2 p K K 2 b 2p 2 b 1p = 1 d d = 1− + d+ D 2d+ D = d D 4 3 ( K ) 2 4 K3 3 K Re 2 b p = + phase 2 busbar 1 or 2 d α = 1 + 3d + D d β = D+ 2d d γ =− 1 + D + d = αa + β + γa K K . e Table 2.7 Symmetrical associated-phase layout Amplification Coefficients 2 = jφ ϕ u = 0 ϕ u = 0 ϕ u = 0 2π φ ν 3 2 2 π − ± where ν is integer remarks calculated by IEC 60865 or by advanced method. 2 b K2 p < 1 if Imono = Itri
d)Methods used IEC 60865 method : - Calculation of amplification factor, - IEC Calculation on the reference case and application of amplification factor to : * force on support of rigid conductors (Fd), * resulting conductor stress (σtot). Advanced method : The amplification factor applies, for the calculation, on the same bases (τ and te) between the reference case and the considered situation ( for example a three-phase fault on two busbars). Designers have to determine the bar and the phase for which the stress is maximum. e) Examples Separate Asymmetrical Symmetrical phase (1) phase (2) phase (3) d (m) 8.8 2 2 D (m) 4.8 2 2 α -1.186 1.5 0.2 β 1 1.5 -0.750 γ -0.254 0.333 -0.167 φ -25.1° 180° 37.5° ||K|| 1.9 1.167 0.83 ϕmax 132.57° 60° 161.2° ϕmin 42.57° -30° 71.2° K 1.8102 1.5556 0.9921 ϕmax Aster (4) 132.6° 60° 165° Kaster (4) 1.78 1.5552 0.9477 % 1.67 0.02 4.47 Table 2.8 Example of caculation for various structure types, for three-phases fault on two busbars. (1) : The maximum response is obtained for a three-phases fault on two busbars on phase 2 bar 1. (2) : As for (1), the maximum response is for a three-phases fault on two busbars on phase 3 bar 1. (3) : In the case of phase 1 bar 1 and Imono = Itri, d 3 for < , the maximum response is obtained D 2 for a three-phase fault on one busbar. For d 3 > , the phase to earth fault on two busbars D 2 has the maximum response. (4) : Results obtained by the Code ASTER. Given a fault flowing on two busbars, the Code ASTER of E.D.F. allows one to compute the influence of the phase variation over the force applied to the central bar. The results have been noted on the Figure 2.27, Figure 2.28 and Figure 2.29. 35 Figure 2.27 Separate-phase layout Force on central bar during a three-phase fault on two busbars. Figure 2.28 Asymmetrical associated phase layout Maximum force during a three-phase fault on two busbars Figure 2.29 Symmetrical associated phase layout Force on outside bar during a three-phase fault on two busbars.
Page 1 and 2: THE MECHANICAL EFFECTS OF SHORT-CIR
Page 3 and 4: 3.7. Special problems .............
Page 5 and 6: 1. INTRODUCTION 1.1. GENERAL PRESEN
Page 7 and 8: The pinch (first maximum) can be ve
Page 9 and 10: 1.3.3.2 Supporting structures Simil
Page 11 and 12: This method allows to state the sho
Page 13 and 14: and becomes the static force in equ
Page 15 and 16: (2.26) M el-pl ⎡ y d/ 2 y = 2⎢
Page 17 and 18: V σ 3 2,5 2 1,5 1 mechanical reson
Page 19 and 20: V F 3 2,5 2 1,5 1 mechanical resona
Page 21 and 22: 1. eigenmode 2. eigenmode 3. eigenm
Page 23 and 24: (2.43) U E max max = 1 2 1 = 2 l
Page 25 and 26: m (2.47) M = σm = Z mσ m dm 2 J w
Page 27 and 28: profiles in Figure 2.14 it follows
Page 29 and 30: (2.71) 2 tube tube 4 tube U ∂U
Page 31 and 32: 2.3.2. Influence of two busbars a)
Page 33: Asymmetrical associated-phase layou
Page 37 and 38: Section 3.4 deals with the effects
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
Page 45 and 46: This value has to be compared with
Page 47 and 48: EN 60865-1 [Ref 3] with the assumpt
Page 49 and 50: a) b) c) 3 m 2 1 0 -1 δ m δ 1 Mov
Page 51 and 52: Droppers with fixed upper ends Duri
Page 53 and 54: Manuzio developed a simplified meth
Page 55 and 56: From prior tests the stiffness valu
Page 57 and 58: Force / kN Force / kN 100mm 200mm 4
Page 59 and 60: 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 ESL F
Page 61 and 62: 3.7. SPECIAL PROBLEMS 3.7.1. Auto-r
Page 63 and 64: for the maximum outswing and the ma
Page 65 and 66: Zug-/Druck-Kraft in kN Z eit in s F
Page 67 and 68: The numerical resolution of these e
Page 69 and 70: 4. GUIDELINES FOR DESIGN AND UPRATI
Page 71 and 72: standards and tested under specifie
Page 73 and 74: 5. PROBABILISTIC APPROACH TO SHORT-
Page 75: f G f(L) Lt Ls 75 G(L) ∞ R = ∫
Page 78 and 79: Only the lowest break values are im
Page 80 and 81: 5.2.2. A global Approach For analyz
Page 82 and 83: to multinode operation can compared
Page 84 and 85:
5.2.3.1.7. Proposed Approach The st
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The first case (Figure 5.14) corres
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centers are normally balanced by ba
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Remark: The risk integral (5.3) can
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1,0 0,8 0,6 0,4 0,2 0,0 -0,2 -0,4 -
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The histograms below for the instan
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5.2.3.8.2. Reliability based design
Page 98 and 99:
power converter in a three phase br
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The time functions of the currents
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a) Determination of the linear mome
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Figure 6.11 Coefficients mJs1 and m
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7. REFERENCES Ref 1. IEC TC 73/CIGR
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Ref 42. Stein, N.; Miri, A.M.; Meye
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Ref 80. Fiabilité des structures d
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8. ANNEX 8.1. THE EQUIVALENT STATIC
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Figure 8.2 What is ESL in this case
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As the eigenfrequency of the portal
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Figure 8.9 Case 3 (third eigenfrequ
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Both cases have the same 200-kV-arr
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a) c) 5 +25 % 0 % 4 3 F c / F st 2
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a) b) 24 kN 20 +25 % 0 % 2,0 m F c
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Case *11 (EDF 1990) This is the 102
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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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