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- High Voltage Engineering - HVE
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**Jawaharlal Nehru Technological University Anantapur (JNTU) College of Engineering (CEP), Pulivendula, Pulivendula, Andhra Pradesh, India - JNTUACEP**- 6 Topics
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Page-2

- introduction to high voltage technology - ( 2 - 13 )
- break down in gaseous and liquid dielectrics - ( 14 - 41 )
- breakdown in solid dielectrics - ( 42 - 61 )
- generation of high D.C and A.C voltages and currents - ( 62 - 120 )
- over voltage phenomenon and insulation co-ordination - ( 121 - 178 )
- high voltage testing of eletrical apparatus - ( 179 - 196 )

Topic:

UNIT-I INTRODUCTION TO HIGH VOLTAGE TECHNOLOGY AND APPLICATIONS INTRODUCTION The potential at a point plays an important role in obtaining any information regarding the electrostatic field at that point. The electric field intensity can be obtained from the potential by gradient operation on the potential i.e. E = – ∇ V ...(1) which is nothing but differentiation and the electric field intensity can be used to find electric flux density using the relation D = εE ...(2) The divergence of this flux density which is again a differentiation results in volume charge density. ∇ . D = ρv ...(3) Therefore, our objective should be to evaluate potential which of course can be found in terms of, charge configuration. However it is not a simple job as the exact distribution of charges for a particular potential at a point is not readily available. Writing εE = D in equation (3) we have εE = ρv or or – ∇ ε ∇ V = ρv 2 ε ∇ V = – ρv ρv 2 ∇ V=– ε ...(4) This is known as Poisson‘s equation. However, in most of the high voltage equipments, space charges are not present and hence ρv = 0 and hence equation (4) is written as 2 ∇ V=0 ...(5) Equation (5) is known as Laplace‘s equation or If ρv = 0, it indicates zero volume charge density but it allows point charges, line charge, ring charge and surface charge density to exist at singular location as sources of the field. Here ∇ is a vector operator and is termed as del operator and expressed mathematically in cartesian coordinates as ∇= ∂ ax ∂ ay ∂ az ∂x ∂y ∂z where a x , ay and az are unit vectors in the respective increasing directions. ...(6)

Hence Laplace‘s equation in cartesian coordinates is given as 2 2 2 ∂ V ∂ V ∂ V 2 ∇ V = ∂x 2 ∂y 2 ∂z 2 = 0 ...(7) Since ∇ . ∇ is a dot produce of two vectors, it is a scalar quantity. Following methods are normally used for determination of the potential distribution (i) Numerical methods (ii) Electrolytic tank method. Some of the numerical methods used are (a) Finite difference method (FDM) (b) Finite element method (FEM) (c) Charge simulation method (CSM) (d) Surface charge simulation method (SCSM). FINITE DIFFERENCE METHOD Let us assume that voltage variations is a two dimensional problem i.e. it varies in x-y plane and it does not vary along z-co-ordinate and let us divide the interior of a cross section of the region where the potential distribution is required into squares of length h on a side as shown in Fig. 0.1. y V2 b V3 c a V0 V1 x d V 4 Fig. 0.1 A portion of a region containing a two-dimensional potential field divided into square of side h . Assuming the region to be charge free ∇ . D = 0 or ∇ . E = 0 and for a two-dimensional situation ∂Ex ∂E y =0 ∂x ∂y and from equation (7) the Laplace equation is 2 2 ∂ V ∂ V 2 2 ∂x ∂y =0 ...(8) Approximate values for these partial derivatives may be obtained in terms of the assumed values (Here V0 is to be obtained when V1, V2, V3 and V4 are known Fig. 1.

∂V ∂x a V1 − V0 h ∂V ∂x and c V0 − V3 h ...(9) From the gradients ∂ V 2 ∂x ∂x a c V1 − V0 −2 V0 V3 h 0 h 0 V2 − V0 − V0 V4 2 h 2 ∂ V 2 ∂y Similarly − ∂V ∂V ∂x 2 ...(10) Substituting in equation (8) we have 2 ∂ V ∂x 2 ∂ 2 V V1 V2 V3 V4 − 4V0 = 0 ∂y 2 h2 1 (V1 + V2 + V3 + V4) ...(11) 4 As mentioned earlier the potentials at four corners of the square are either known through computations or at start, these correspond to boundary potentials which are known a priori. From equation (11) it is clear that the potential at point O is the average of the potential at the four neighbouring points. The iterative method uses equation (11) to determine the potential at the corner of every square sub-division in turn and then the process is repeated over the entire region until the difference in values is or V0 = less than a prespecified value. The method is found suitable only for two dimensional symmetrical field where a direct solution is possible. In order to work for irregular three dimensional field so that these nodes are fixed upon boundaries, becomes extremely difficult. Also to solve for such fields as very large number of V(x, y) values of potential are required which needs very large computer memory and computation time and hence this method is normally not recommended for a solution of such electrostatic problems. FINITE ELEMENT METHOD This method is not based on seeking the direct solution of Laplace equation as in case of FDM, instead in Finite element method use is made of the fact that in an electrostatic field the total energy enclosed in the whole field region acquires a minimum value. This means that this voltage distribution under given conditions of electrode surface should make the enclosed energy function to be a minimum for a given dielectric volume v. We know that electrostatic energy stored per unit volume is given as 2 W= 1 ∈E ...(12) 2 For a situation where electric field is not uniform, and if it can be assumed uniform for a differential volume δv, the electric energy over the complete volume is given as 1 W= 2 z V 1 2 ∈ ( − ∇V ) dv ...(13)

To obtain voltage distribution, our performance index is to minimise W as given in equation (13). Let us assume an isotropic dielectric medium and an electrostatic field without any space charge. The potential V would be determined by the boundaries formed by the metal electrode surfaces. Equation (13) can be rewritten in cartesian co-ordinates as W= 1 2 L ∈ zzz F ∂V MG H M I 2 F ∂V I 2 F ∂V I J ∂x K G J ∂y K H G H 2 J ∂z K N O ...(14) P dxdydz P Q Assuming that potential distribution is only two-dimensional and there is no change in potential ∂V along z-direction, then ∂z = 0 and hence equation (14) reduces to WA = z L M 1∈ zz M 2 N R ∂V I F ∂V I UO 2 2 P dxdy G J H ∂x K S | ...(15) | |F G J ∂y K H T V |P WQ Here z is constant and WA represents the energy density per unit area and the quantity within integral sign represents differential energy per elementary area dA = dxdy. In this method also the field between electrodes is divided into discrete elements as in FDM. The shape of these elements is chosen to be triangular for two dimensional representation and tetrahedron for three dimensional field representation Fig. 0.2 (a) and (b). Vk Vk V Vj h V V i j Vi Fig. (a) Triangular finite element (b) Tetrahedron finite element. The shape and size of these finite elements is suitably chosen and these are irregularly distributed within the field. It is to be noted that wherever within the medium higher electric stresses are expected e.g. corners and edges of electrodes, triangles of smaller size should be chosen. Let us consider an element e1 as shown in Fig. 0.2(a) as part of the total field having nodes i, j and k in anti-clockwise direction. There will be a large no. of such elements e1, e2 .....eN . Having obtained the potential of the nodes of these elements, the potential distribution within each elements is required to be obtained. For this normally a linear relations of V on x and y is assumed and hence the first order approximation gives V(x, y) = a1 + a2x + a3 y ...(16)

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