HIGH VOLTAGE TECHNIQES Basic Electrode Systems (3) Assistant Professor Suna BOLAT KRÖGER Eastern Mediterranean niversity Department of Electric & Electronic Engineering 1
Basic electrode systems Different configurations Parallel plate electrodes Co-axial cylinders Concentric spheres 2
Cylindrical Electrode systems r 2 : applied voltage r 1, r 2 : radii of the spheres r 1 Co-axial Cylinders 3
Electric field Which coordinate system we should use? Cylindrical coordinates!!! What does it depend on? E (hence V) only depends on the r-coordinate!!! r,, z: cylindrical coordinates 4
Cylindrical coordinate system 5
Electric field and potential equations Electrical charges We can use Gauss Law Laplace s equation In cylindrical coordinates 6
By Laplace s equation Laplace s equation in cylindrical coordinates V = 2 V r 2 + 1 r V r + 1 2 V r 2 θ 2 + 2 V z 2 = 0 (dependent variable is r) 0 0 V = d2 V dr 2 + 1 r dv dr = 0 V = V(r) 2 nd degree differential equation 7
ln d dv dr dr d dv dr dv dv dr + 1 r dv dr = 0 = 1 r dr dr = ln r + A = ln A r dv dr = A r dv = A r dr 8
Solution for this differential equation General solution: V = V r = A ln r + B Constants A and B is determined by the data of the problem which is called boundary conditions. 9
Boundary conditions r = r 2 V = V 2 = 0 0 = A ln r 2 + B B = A ln r 2 r = r 1 V = V 1 = = A ln r 1 + B = A ln r 1 A ln r 2 = A ln r 1 A = ln r 1 r2 r 2 A = ln r & B = 2 ln r 2 ln r 2 10
A = ln r & B = 2 ln r 2 ln r 2 Lets put those constants into the general solution: V = A lnr + B = ln r ln r + 2 ln r ln r 2 2 = ln r ln r + ln r 2 2 11
Solution Potential equation for the co-axial cylinders V = V(r) = ln r ln 2 r 2 r 12
Potential change by r 13
Solution Electric Field Equation for co-axial cylinders E = dv dr = ln r 2 1 r E changes inversely proportional to r 14
Equipotential lines Equipotential lines 15
Electric field lines Equipotential lines Electric field lines 16
Electric field lines ALWAYS perpendicular to equipotential lines (ORTHOGONAL) 17
For r = r 1 For r = r 2 E = E r 1 = E = E r 2 = r 1 ln r 2 r 2 ln r 2 = E max = E min 18
Electric field change by r Similar to concentric spheres. What s the difference? 19
When does it breakdown? Emax Eb Discharge in the insulation!!! What is the difference of breakdown condition from that parallel plate configuration?? 20
Electric field equation by electrical charges Gauss s Law: DdS = Q For uniform symmetrical surfaces: D. S = Q (S: surface of the electrode) εe. S = Q sing these = E = Q εs E dr Can you find electric potential and field equations? 21
Capacitance r 2 r 1 Co-axial cylindrical capacitor S: area of the crosssectional surface 22
Capacitance Q = C. charge capacitance voltage C = Q = D. S εe. S = 1 E = ln r & S = 2πrl 2 r ε ln r 1 2 r 2πrl C = 23
Capacitance F/m m C = 2πε l 1 ln r 2 F : Dielectric constant of the insulation = 0 r [F/m] r : relative dielectric constant (no unit), relative permittivity 0 = 8.854 10-12 F/m dielectric constant for space, permittivity r 1 : radius of the inner electrode [m] r 2 : radius of the outer electrode [m] cm cm 24
Multilayer dielectric co-axial cylinder This dielectric configuration is connected in SERIES! r 1 r 2 r 3 R C 1 C 2 C 3 l 25
This dielectric configuration is connected in SERIES! I = I 1 = I 2 = I 3 = = I n Q = Q 1 = Q 2 = Q 3 = = Q n C = C 1 1 = C 2 2 = C 3 3 = = C n n is the voltage applied to the entire system 1, 2,, n are voltages across the corresponding dielectric layers = 1 + 2 + 3 + + n 26
Voltages across each dielectric layer: C = C 1 1 = C 2 2 = C 3 3 = = C n n 1 = 2 = C C 1 C C 2 n = C C n 27
Equivalent capacitance of the system X C = X C1 + X C2 + X C3 + + X Cn 1 ωc = 1 + 1 + 1 + + 1 ωc 1 ωc 2 ωc 3 ωc n 1 C = 1 C 1 + 1 C 2 + 1 C 3 + + 1 C n C : equivalent capacitance of the system, total capacitance of the system : voltage applied to the entire system C i : capacitance of the i th dielectric layer i : voltage across i th dielectric layer 28
Calculating capacitance C 1 = 2πε 1 l 1 C = 1 2πε 1 l 1 ln r 2, C 2 = 2πε 2 l 1 ln r 2 + 2πε 2 l 1 1 ln r 3 r2 1 ln r 3 r2,, C n = 2πε n l + + 2πε n l 1 1 ln R r n 1 ln R r n = 1 2πl 1 ε 1 ln r 2 r 1 + 1 ε 2 ln r 3 r 2 + + 1 ε n ln R r n = 1 2πl n i=1 1 ε i ln r i+1 r i A 29
Total capacitance, equivalent capacitance A = n i=1 1 ε i ln r i+1 r i = 1 ε 1 ln r 2 r 1 + 1 ε 2 ln r 3 r 2 + + 1 ε n ln R r n 1 C = A 2πl C = 2πl A 30
Potential and electric field across 1 st dielectric layer 1 = C C 1 = 2πl A 1 2πε 1 l ln r 2 = Aε 1 ln r 2 r 1 E 1max = 1 r 1 ln r 2 = E 1min = 1 r 2 ln r 2 = ln r 2 Aε 1 r 1 ln r 2 = ln r 2 Aε 1 r 2 ln r 2 = Aε 1 r 1 Aε 1 r 2 We can write the same equations with respect to applied voltage () 31
Potential and electric field across i th dielectric layer i = C. C i = A. ε i. l i. ln r i+1 r i E imax = E imin = Voltage across i th layer i r i. ln r = i+1 r i i r i+1. ln r i+1 r i = A. ε i. r i A. ε i. r i+1 Applied Voltage 32
Electric field strength across any dielectric layer, E i,max, becomes greater than breakdown strength of that dielectric, there will be an electrical discharge. E i,max E bi Discharge in that layer of the insulation!!! Breakdown, short circuit in i th dielectric layer 33
Electric field distribution E E max1 E min1 r 1 r 2 r 3 r 4 R r 34
niform stress condition E E max r 1 r 2 r 3 R If l 1 = l 2 = = l n = l = constant r E 1max = E 2max = = E nmax 35
A. ε 1. r 1. l = A. ε 2. r 2. l = = A. ε n. r n. l ε 1. r 1 = ε 2. r 2 = = ε n. r n The system is uniformly stressed! 36
Remarks What are the applications for co-axial cylinders that you can think of? What does a cable look like? 37
Applications for co-axial cylinders Cables Single Phase Coaxial cable 38
Bushings 39
Bushing with layers of different permittivity 40
Layers are connected in series Q = Q 1 = Q 2 = = Q n C = C 1 1 = C 2 2 = = C n n C 1 = 2πε 1l 1 l n r 2, C 2 = 2πε 2 l 2 r C l 3 n = n r2 C i = 2πε il i l n r i+1 r i 2πε n l n l n R r n 41
1 C = 1 C 1 + 1 C 2 + + 1 C n 1 C = 1 2π n i=1 1 ε i l i ln r i+1 r i = A 2π C = 2π A, A = 1 ε i l i ln r i+1 n i=1 r i 42
1 = C. C 1 = 2π A. 2πε 1. l 1 ln r 2 = Aε 1 l 1 ln r 2 r 1 i = C. C i = A. ε i. l i. ln r i+1 r i Voltage across each layer 43
E 1max = 1 r 1. ln r 2 = E 1min = 1 r 2. ln r 2 =. ln r 2 A. ε 1. l 1 r 1. ln r = 2. ln r 2 A. ε 1. l 1 r 2. ln r = 2 A. ε 1. l 1. r 1 A. ε 1. l 1. r 1 44
In general: Voltage across i th layer Applied Voltage E imax = E imin = i r i. ln r = i+1 r i i r i+1. ln r i+1 r i = A. ε i. l i. r i A. ε i. l i. r i+1 45
Potential and electric field across i th dielectric layer i = C C i = 2πl A 2πε i l ln r i+1 r i = Aε i ln r i+1 r i E imax = i r i ln r i+1 r i E imin = i r i+1 ln r i+1 r i Write down voltage equations for each layer into field equations. See what happens!!! Can you draw a graph for field distribution on the surface?
E imax E bi Discharge! To make the field disribution more uniform; (uniformly stressed) E 1max = E 2max = = E nmax A. ε 1. l 1. r 1 = A. ε 2. l 2. r 2 = = A. ε n. l n. r n ε 1. l 1. r 1 = ε 2. l 2. r 2 = = ε n. l n. r n 47
Question How do you think we can find the total capacitance of this configuration? HINT: how the capacitors are connected? 48
Next topic... Breakdown behaviour in co-axial cylinders. 49