The mechanical effects of short-circuit currents in - Montefiore

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The time functions of the electromagnetic forces consist of four partial functions: (2.13) F( t) = F 10 + F 422 43 ω ( t) + Fg ( t) + Fω ( t) 1 44243 4 steady state decaying – constant term F0, arithmetic mean of F(t) in steady state; – undamped oscillation F2ω(t) at double the electrical frequency; – exponential term Fg(t), decaying with τ/2; – oscillation F2ω(t) with electrical frequency, decaying with τ. τ is the time constant of the electrical network. The maximum values are: – three-phase short-circuit, inner conductor L2: µ 0 3 2 l µ 0 2 l (2.14) Fm3, L2 = ip3 = 0, 866 ip3 2π 2 am 2π am – three-phase short-circuit, outer conductors L1 and L3: (2.15) F m3, L1 = F = m3, L3 0, 808 µ 0 3 + 2 = 2π 8 µ 0 i 2π – line-to-line short-circuit: (2.16) F m2 µ 0 = i 2π = 0, 750 2 p2 l a m µ 0 i 2π 2 p3 2 p3 m l a m µ ⎛ 0 = ⎜ 2π ⎜ ⎝ l a 3 2 3 i i p3 2 p3 ip3 and ip2 are the peak short-circuit currents during three-phase respective line to-line short-circuits. They can be calculated from the initial short-circuit current I k′′ : i p = κ 2I k′′ [Ref 11, Ref 12]. The time patterns of the forces and the investigations of the partial functions drive to the following comments: a) Comparison of the forces at the inner conductor L2 with those at the outer conductors L1, L3 during three-phase short-circuit: – The constant term F0 is of considerable magnitude at L1 and L3 and not present for L2. – The maxima of steady-state oscillation F2ω(t) are twice as large for L2 as they are for L1, L3. The maxima of Fω and Fg are somewhat larger for L2 than for L1, L3. – The time histories for the resultant forces show a noticeable difference, however the maxima at L2 are always about 7 % higher than at L1, L3 for all R/X, both at transient and steady state. Considering only the forces it cannot be said which conductor will be stressed higher: L1 and L3, or L2, although the maximum is greater at L2 than at L1, ⎞ ⎟ ⎠ 2 l a m l a m 12 L3. Section 2.2.3 describes that with HVarrangements with conductor frequency much lower than the system frequency the stresses on conductors L1, L3 are higher than L2, where as for MV- and LV-arrangements with frequencies equal or greater than system frequency the stress in L2 is decisive. b) Comparison of the forces during line-to-line shortcircuit with those at the outer conductors L1, L3 during three-phase short-circuit: – The constant terms F0 are equal. – The maxima of all partial functions are somewhat smaller in the case of a line-to-line short-circuit. – The time histories of the resultant forces are very similar. In the case of a line-to-line short-circuit, the maxima are 7 % smaller. The line-to-line short-circuit stress is somewhat lower than three-phase short-circuit; Therefore the line-to-line short-circuit needs no additional investigation. The maximum force has the highest value at the inner conductor L2 during three-phase short-circuit and is used as static force in equation (2.2) and in the standard. The static force during a line-to-line short-circuit is equation (2.3). In the case of conductors not having circular or tubular profiles, the effective distances am of the main conductors differ from the center-line distances. am can be calculated according to the standard; further information is given in [Ref 4, Ref 13, Ref 14]. The influence of the current displacement in the conductors can be calculated with Current-Splitting- Method [Ref 15]. As shown in [Ref 16], it is of low importance; therefore equations (2.2) and (2.3) are valid with sufficient accuracy. The stress due to the forces between the sub-conductors is calculated separately and superposed according to section 2.2.6. In each of the sub-conductors, the current iL/n flows in the same direction. The greatest force acts on the outer sub-conductors where the forces due to the n - 1 sub-conductors are added: 2 µ 0 F1 = 2π (2.17) ⎛ iL ⎞ ⎜ ⎟ ⎝ n ⎠ ⎡ 1 ⎢ ⎢⎣ as,12 1 + as,13 1 + L+ a µ 0 = 2π 2 ⎛ iL ⎞ ls ⎜ ⎟ ⎝ n ⎠ a s s,1n ⎤ ⎥ l ⎥⎦ ls is the distance between two spacers and as,1i the effective distance between conductors 1 and i. The term in brackets can be summarised in the effective distance as. The maximum value is equivalent to those of the line-to-line short-circuit according to equation (2.16): (2.18) µ 0 ⎛ ip ⎞ ls Fs = ⎜ ⎟ 2π ⎜ n ⎟ ⎝ ⎠ as 2 s
and becomes the static force in equation (2.4). 2.2.3. Bending stresses in the conductors taking into account plastic effects and forces on the substructures The determination of the short-circuit strength according to the rules of statics and under the assumption that deformations occur in the elastic range leads to a design of the busbar which requires less supporting distances, higher cross-sections and /or more robust supports, meaning higher costs. If small permanent plastic deformations are permitted after short-circuit a favourable design is obtained. Therefore the verification of safe loading is done taking advantage of plastic load capacity. The calculation is done according to the theory of plastic hinges similar to the one done for steel structures [Ref 17, Ref 18]. 2.2.3.1 Short-circuit strength of conductors The static load Fm cause in a beam a static stress (2.19) σ m, stat M = Z pl, max m Fml = β 8Z with the moment Mpl,max according to Table 2.1. With different factor β the same equation can be used for different types and numbers of supports. The dynamic response of the structure is considered by the use of the factors Vσ and Vr [Ref 1, Ref 4, Ref 8, Ref 9, Ref 19, Ref 20]. Hence the rating value is as follows: (2.20) σ m m F l = VσVr σm, stat = βVσVr 8Z m m Sub-conductors are fixed by connecting pieces in the span. The outer sub-conductors oscillate towards each other and can be handled as a single span beam fixed in the connecting pieces. According to the equations (2.19) and (2.20) the structure response due to the forces between the sub-conductors leads to the rating stress in the sub-conductor: (2.21) M pl, max F l σ s = Vσ sVrσ s, stat = VσsVr = VσsVr Z 16Z s s s Conductors are assumed to withstand a short-circuit if they do not show a remarkable permanent deformation [Ref 4, Ref 21]. In the case of short actions this is true if the conductor is not stressed with higher stress than with twice the stress corresponding to the yielding point of the material taking advantage of its plasticity: = 2R (2.22) zul p0,2 σ Tests show the occurrence of a permanent deformation of 0,3 ... 0,5 % of the supporting distance which is 13 equivalent to a conductor lengthening of about 0,03 % 1 . In equation (2.22), the factor 2 of plasticity arises when a beam fixed at both ends has plastic deformations within the span and in addition there is buckling in the rigid supporting points; it changes from a fix support to a partial one. In steel construction the maximum capacity is reached when a plastic hinges are used. This means that the stress at the fix points and in the span is at the yielding point and there is full plasticity. The factor β considers the plastic hinges at the rigid support point and the factor q the plastic hinge in the span. When going from complete to partial fixation the maximum moment Mpl,max occurs in the fixation before the maximum elastic moment Mel,max is reached. By this the maximum moment at the fixation is decreased. Both moments are given in Table 2.1 and the comparison show that a greater loading is possible. In the case of a beam fixed at both ends 33 % can be gained: M el, max Fml / 12 4 (2.23) = = = 1, 33 M pl, max Fml / 16 3 The single span beam supported at both ends has no plastic hinges in the supports. The spans with continuous beams are calculated separately and the moments are estimated. Arrangements with two spans are equivalent to a beam supported/fixed. With three and more spans the outer spans are nearly equivalent to a beam supported/fixed and the inner spans to a beam fixed/fixed; the moments are greater in the outer spans than in the inner ones, which can be found by comparison with a single span: The outer spans are decisive. The plastic moments for different types of beams and supports are related to the moment in a beam supported at both ends. Hence the factor β is calculated, which is also given in Table 2.1: (2.24) β = M M pl, max el, max, supported/ supported The plastic behaviour in the span can be explained best with materials having a distinctive yielding point. This is not existent neither for aluminum nor for copper, as shown in Figure 2.1a. It is substituted by an ideal elastic-plastic characteristic. The following is shown in the case of an rectangular profile, which is easier to demonstrate and is analogous to other profiles. 1 The tests have been performed in 1942 and 1944 by Siemens-Schuckert factory in Berlin
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: This method allows to state the sho
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 and 34: Asymmetrical associated-phase layou
Page 35 and 36: d)Methods used IEC 60865 method : -
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
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for the maximum outswing and the ma
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Zug-/Druck-Kraft in kN Z eit in s F
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The numerical resolution of these e
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4. GUIDELINES FOR DESIGN AND UPRATI
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standards and tested under specifie
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5. PROBABILISTIC APPROACH TO SHORT-
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f G f(L) Lt Ls 75 G(L) ∞ R = ∫
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Only the lowest break values are im
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5.2.2. A global Approach For analyz
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to multinode operation can compared
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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
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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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The mechanical effects of short-circuit currents in - Montefiore

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