transformer engineering design and practice 1_phần 2

Appendix B: Stress and Capacitance Formulae In this appendix, formulae are derived for electric stress and capacitance for commonly existing electrode configurations in transformers such as two round electrodes or round electrode and plane. B1 Stress Calculations The information about the electric field intensity and potential field between two parallel cylindrical electrodes can be found by considering the respective equivalent line charges. Consider two line charges +ρL and -ρL (charges per unit length) placed at x=+m and x=-m respectively as shown in figure B1. Now, due to single line charge ρL, the electric field intensity at a distance r is given by (B1) where e is permittivity of medium. The potential reckoned from a distance R is (B2) The resultant potential at point A (figure B1) due to line charges +ρL and-ρL is (B3) 459 Copyright © 2004 by Marcel Dekker, Inc. 460 Appendix B Figure B1Two line charges placed at x=-m and x=+m Let us now find the nature of equipotential surface having potential of u. From equation B3 we get But from figure B1 we have Solving by componendo and dividendo, By algebraic manipulations we get Copyright © 2004 by Marcel Dekker, Inc. Appendix B 461 (B4) This is the equation of a circle with radius and center Thus, the equipotential surface is a cylinder which intersects the x-y plane in a circle with radius r and center at (s, 0). From the above expressions for radius and center we get (B5) (B6) By substituting the value of m in the equation for radius we have (B7) Now, from equations B5 and B7 we get Thus, we get the expression for potential as (B8) Now, we will consider two parallel cylindrical conductors of radii R1 and R2, placed such that the distance between their centers is 2s. The electric field intensity and potential between the two conductors are calculated by considering the corresponding two equivalent line charges as shown in figure B2. Copyright © 2004 by Marcel Dekker, Inc. 462 Appendix B Figure B2 Configuration of two parallel cylindrical conductors Now, S1+S2=2s (B9) Using equation B6 we can write (B10) By solving equations B9 and B10 we get The electric field intensity at point P on the surface of the conductor on the right side is given by Now, by putting the value of in the above equation we get Copyright © 2004 by Marcel Dekker, Inc. Appendix B 463 By putting the value of s1 obtained earlier in the above equation we get (B11) Now, by using equation B8 for potential, the potential difference between points P and Q is given as By putting the values of s1 and s2 in the above equation and simplifying, Putting this value in equation for Ep (equation B11) we have (B12) where Copyright © 2004 by Marcel Dekker, Inc. ... - tailieumienphi.vn