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High Voltage Engineering

by Jntu Heroes
Type: NoteInstitute: Jawaharlal nehru technological university anantapur college of engineering Offline Downloads: 218Views: 4709Uploaded: 10 months agoAdd to Favourite

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Jntu Heroes
LECTURE NOTES ON HIGH VOLTAGE ENGINEERING IV B. Tech I semester (JNTUH-R09) ELECTRICAL AND ELECTRONICS ENGINEERING 1
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. 2 ...(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. 3
∂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 4 ...(13)

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