# Chapter 4: Breakdown in Soild Dielectrics¶

## Example 1: pg 123¶

In :
#example 4.1
#calculation of heat generated in specimen due to dielectric loss

#given data
epsilonr=4.2#value of the dielectric constant
tandelta=0.001#value of tandelta
f=50#value of frequency(in Hz)
E=50*10**3#value of electric field(in V/cm)

#calculation
#from equation of dielectric heat loss......H=(E*E*f*epsilonr*tandelta)/(1.8*10^12)
H=(E*E*f*epsilonr*tandelta)/(1.8*10**12)
#results
print 'The heat generated in specimen due to dielectric loss is (mW/cm^3) = ',round(H*10**3,3)

The heat generated in specimen due to dielectric loss is (mW/cm^3) =  0.292


## Example 2: pg 123¶

In :
#example 4.2
#calculation of voltage at which an internal discharge can occur

#given data
d1=1#thickness(in mm) of the internal void
dt=10#thickness(in mm) of the specimen
epsilon0=8.89*10**-12#electrical permittivity(in F/m) of free space
epsilonr=4#relative permittivity of the dielectric
Vb=3#breakdown strength(in kV/mm) of air

#calculation
d2=dt-d1
epsilon1=epsilon0*epsilonr#electrical permittivity(in F/m) of the dielectric
V1=Vb*d1#voltage at which air void of d1 thickness breaks
V=(V1*(d1+(epsilon0*d2/epsilon1))/d1)
#results
print 'the voltage at which an internal discharge can occur is (kV.) = ',round(V,2)

the voltage at which an internal discharge can occur is (kV.) =  9.75


## Example 3: pg 124¶

In :
#example 4.3
#calculation of the dimensions of electrodes in coaxial cylindrical capacitor
from math import pi,exp
#given data
epsilon0=(36*pi*10**9)**-1#electrical permittivity(in F/m) of free space
#consider high density polyethylene as the dielectric material
epsilonr=2.3#relative permittivity of high density polyethylene
l=0.2#effective length(in m)
C=1000*10**-12#capacitance(in F) of the capacitor
V=15#operating voltage(in kV)
Emax=50#maximum stress(in kV/cm) for breakdown stress 200 kV/cm and factor of safety of 4

#calculation
#from equation of capacitance of coaxial cylindrical capacitor
#C=(2*%pi*epsilon0*epsilonr*l)/(lod(d2/d1)).............(1)
#from equation of Emax occuring near electrodes
#Emax=V/(r1*(log(r2/r1)))...............................(2)
#from equation (1) and equation (2),we get
logr2byr1=(2*pi*epsilon0*epsilonr*l)/C#logd2/d1 = logr2/r1
r1=V/(Emax*logr2byr1)#from equation (1)
r2=r1*exp(logr2byr1)
#results
print 'the value of inner diameter of electrodes in coaxial cylindrical capacitor is (cm) = ',round(r1,2)
print '\nthe value of outer diameter of electrodes in coaxial cylindrical capacitor is (cm) = ',round(r2,2)
print '\nthe thickness of the insulation is (mm) = ',round((r2-r1)*10,1)

the value of inner diameter of electrodes in coaxial cylindrical capacitor is (cm) =  11.74

the value of outer diameter of electrodes in coaxial cylindrical capacitor is (cm) =  12.04

the thickness of the insulation is (mm) =  3.0