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


  1. [1] H. Cookson, P.C. Bolin, H.C. Doepken, R.E. Wootton, C.M.Cooke, & J.G. Trump Recent research in the United States onthe effect of particle contamination reducing the breakdownvoltage in compressed gas insulated system, Int. Conf. OnLarge High Voltage System, Paris, 1976.
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  5. [5] J. Amarnath, B.P. Singh, C. Radhakrishna, & S. Kamakshiah,Determination of particle trajectory in a gas insulated busductpredicted by Monte-Carlo technique, CEIDP, 1999 during17–21 October 1999, Texas, Austin, USA.
  6. [6] T. Takeuchi, T. Yoshizawa, Y. Kuse, & H. Hama, 3-D non-linear transient electromagnetic analysis of short circuit elec-tromagnetic forces in a three-phase enclosure-type gas insu-lated bus, IEEE Transactions on Magnetics, 36 (4), July 2000,1,754–1,757.AppendixGeneral Expression for Electric FieldThe electric field intensity from the surface of the enclo-sure, due to three conductors at a given point acts simul-taneously. Let the particle move to a distance ‘x’ from theinner surface of the enclosure at the point ‘p’ in Fig. 2.From the graph, the variables in the general formulaare:R1 = Distance between the conductor B and particleθ2 = Angle between the vertical axis and R1x = Position of the particle within the enclosureh = Distance between the centre of the conductor andenclosurer = Radius of the conductorFrom Fig. 2 and Δle123R21 = (h − x)2+ k2− 2(h − x)K Cos 150◦(A.1)The final expression for the three-phase field is given by:E = E2x + E2y (A.2)whereEx = 48.64 × 103cos 0◦ 10.125 − x+(cos 120◦+ cos 240◦)cos θ2R1(A.3)Ey = 48.64 × 103sin 0◦ 10.125 − x+(sin 120◦+ sin 240◦)cos θ2R1(A.4)The motion equation is given by:m¨y(t) =π ∈0 l2E(t0)ln2lr− 1× 48.64 × 103×10.125 − x−cos θ2R1sin 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 andmagnetic 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 isgiven as:Q =πε0l2E(t0)ln 2lr − 1E =Vm sin ωt[r0 − y(t)] ln r0riThe Lorentz force is given by:FLF =πε0l2E(t0)ln 2lr − 1×⎡⎣ Vm sin ωt[r0 − y(t)] ln r0ri− ˙yμI2π[r0 − y(t)]⎤⎦

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