G.V.N. Kumar, J. Amarnath, B.P. Singh, and K.D. Srivastava


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  4. [4] J.R. Lahari & A.H. Qureshi, A review of particle contaminated gas breakdown, IEEE Transactions on Electr. Insul., 16, 1981, 388–398.
  5. [5] J. Amarnath, B.P. Singh, C. Radhakrishna, & S. Kamakshiah, Determination of particle trajectory in a gas insulated busduct predicted by Monte-Carlo technique, CEIDP, 1999 during 17–21 October 1999, Texas, Austin, USA.
  6. [6] T. Takeuchi, T. Yoshizawa, Y. Kuse, & H. Hama, 3-D nonlinear transient electromagnetic analysis of short circuit electromagnetic forces in a three-phase enclosure-type gas insulated bus, IEEE Transactions on Magnetics, 36 (4), July 2000, 1,754–1,757. Appendix General Expression for Electric Field The electric field intensity from the surface of the enclosure, due to three conductors at a given point acts simultaneously. Let the particle move to a distance ‘x’ from the inner surface of the enclosure at the point ‘p’ in Fig. 2. From the graph, the variables in the general formula are: R1 = Distance between the conductor B and particle θ2 = Angle between the vertical axis and R1 x = Position of the particle within the enclosure h = Distance between the centre of the conductor and enclosure r = Radius of the conductor From Fig. 2 and Δle 123 R2 1 = (h − x)2 + k2 − 2(h − x)K Cos 150◦ (A.1) The final expression for the three-phase field is given by: E = E2 x + E2 y (A.2) where Ex = 48.64 × 103 cos 0◦ 1 0.125 − x +(cos 120◦ + cos 240◦ ) cos θ2 R1 (A.3) Ey = 48.64 × 103 sin 0◦ 1 0.125 − x +(sin 120◦ + sin 240◦ ) cos θ2 R1 (A.4) The motion equation is given by: m¨y(t) = π ∈0 l2 E(t0) ln 2l r − 1 × 48.64 × 103 × 1 0.125 − x − cos θ2 R1 sin ωt − mg − ˙y(t)πr(6μKd( ˙y) + 2.656(μrgly(t)0.5 ) (A.5) If charge moves in a region, where the electric field and magnetic fields are simultaneously present. The Lorentz force is given by: F = Q(E + (V × B))Newtons (A.6) The velocity is given by V = ˙y m/s. (A.7) The charge acquired by the particle and the lift-off field is given as: Q = πε0l2 E(t0) ln 2l r − 1 E = Vm sin ωt [r0 − y(t)] ln r0 ri The Lorentz force is given by: FLF = πε0l2 E(t0) ln 2l r − 1 × ⎡ ⎣ Vm sin ωt [r0 − y(t)] ln r0 ri − ˙y μI 2π[r0 − y(t)] ⎤ ⎦

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