Calculation of alternating current losses in stacks and coils made of second generation high temperature superconducting tapes for large scale applications
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...The challenge of simulating superconductors with very large width-to-thickness ratio (such as HTS coated conductors) was first tackled by artificially expanding the superconducting layer’s thickness [36] and then, for systems characterized by a large number of tapes, with the homogenization method [37]....
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"Calculation of alternating current ..." refers background in this paper
...Ï.7;H75, (23) where HÜÝ is the length of the edge joining the vertices E and F as presented in FIG. 12. Then, the magnetic field t within the element is given by: t L ÍzÜ*Ü 7 Ü@5 , (24) 17 where *Ü#TAB#is the tangential component of the magnetic field on the Eth edge. Just like in the previous section, it is easy to see that: ∙ L0 (25) and Ï Ht L 1 Δ Í*ÜHÜ 7 Ü@5 G à. (26) Ag...
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"Calculation of alternating current ..." refers background in this paper
...etize the magnetic field L :*ë,ì;. FIG. 12 Triangular edge element For this purpose, the area coordinates :.5,6,7; are considered as follows: .Ý:,U; L 1 2∆ k=Ý E>ÝT E?ÝU o, (19) where ∆ 5 6 :T5:U6 FU7; ET6:U7 FU5; ET7:U5 FU6;; is the surface area of the triangular element and the constant coefficients =Ý, >Ý and ?Ý are given by: =5 LT6...
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"Calculation of alternating current ..." refers background in this paper
...n from Thakur et al.20. Once multiplied by the volume fraction, the equivalent , ,ä:n; dependence is given by: 8 ,,ä:n; L ,Ö , BÁÍÌ É Ç1 §G6+$ ∥+ 6 E|$D|6 $4 Ì Ê , (14) where $4 L42.65#TAB#mT, Ö , L28#TAB#GA/m6, G L0.29515, Ù L0.7, and $ ∥ and $D are respectively, the parallel and perpendicular components of the magnetic flux density with respect to the tape’s surfa...
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408 citations
"Calculation of alternating current ..." refers methods in this paper
...he element is given by: t L ÍzÜ*Ü 7 Ü@5 , (24) 17 where *Ü#TAB#is the tangential component of the magnetic field on the Eth edge. Just like in the previous section, it is easy to see that: ∙ L0 (25) and Ï Ht L 1 Δ Í*ÜHÜ 7 Ü@5 G à. (26) Again, recalling that the tangential components *Ü,E∈<1,2,3, are space constants and using Ampere’s law Ï Ht Lv, it is easy to see that the cur...
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