# Chapter 7 : Dielectric Materials and Insulation¶

## Example 7.1 Page No : 301¶

In :
# Given
NA = 6.023*10**23           # in mol**-1
d = 1.8                     # g/cm3
Mat = 39.95                 # in mol**-1
epsilon_o = 8.85*10**-12    # F/m2
alpha_e = 1.7*10**-40       # F*m2

# Calculations and Results
N = NA*d/Mat                # in cm**-3
N *= 10**6                  # in m**-3
epsilon_r = 1+(N*alpha_e/epsilon_o)
print("Dielectric constant of solid Ar is {0:.4f}".format(epsilon_r))
# using clausius-mossotti equation
epsilon_r = (1+(2*N*alpha_e/(3*epsilon_o)))/(1-(N*alpha_e/(3*epsilon_o)))
print("using clausius-mossotti equation,  Dielectric constant of solid Ar is {0:.4f}".format(epsilon_r))

Dielectric constant of solid Ar is 1.5213
using clausius-mossotti equation,  Dielectric constant of solid Ar is 1.6309


## Example 7.2 Page No : 307¶

In :
import math

# Given
N = 5*10**28                    # in m**-3
e = 1.6*10**-19                 # in coulombs
Z = 4.0
me = 9.1*10**-31                # in Kg
epsilon_o = 8.85*10**-12        # F/m2
epsilon_r = 11.9

# Calculations and Results
# part(a)
alpha_e = (3*epsilon_o/N)*((epsilon_r-1)/(epsilon_r+2))
print("Electronic polarizability in F/m2 {0:.4g}".format(alpha_e))
# part(b)
# let x=E_loc/E
x = (epsilon_r+2)/3.0
print("Local field is a factor of {0:.4f} greater than applied field".format(x))
# part(c)
wo = math.sqrt(Z*e**2/(me*alpha_e))
fo = wo/(2*math.pi)
print("resonant frequency in Hz is {0:.4g}".format(fo))

Electronic polarizability in F/m2 4.164e-40
Local field is a factor of 4.6333 greater than applied field
resonant frequency in Hz is 2.616e+15


## Example 7.3 Page No : 311¶

In :
# Given
# let epsilon=E
Eo = 8.85*10**-12               # in F/m
Ni = 1.43*10**28                # in m**-3
alpha_e_Cs = 3.35*10**-40       # F m2
alpha_e_Cl = 3.40*10**-40       # F m2
alpha_i = 6*10**-40             # F m2

# Calculations and Results
# (Er-1)/(Er+2)=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-)+Ni*alpha_i)
# let x=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-)+Ni*alpha_i)
# after few mathematical steps we get
# Er=(2*x+1)/(1-x)
x = (1.0/(3*Eo))*(Ni*alpha_e_Cs+Ni*alpha_e_Cl+Ni*alpha_i)
Er = (2*x+1)/(1-x)
print("Dielectric constant at low frequency is")
# similarly
# let y=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-))
# after few mathematical steps we get
# Erop=(2*x+1)/(1-x)
y = (1.0/(3*Eo))*(Ni*alpha_e_Cs+Ni*alpha_e_Cl)
Erop = (2*y+1)/(1-y)
print("Dielectric constant at optical frequency is {0:.4f}".format(Erop))

Dielectric constant at low frequency is
Dielectric constant at optical frequency is 2.7137


## Example 7.6 Page No : 315¶

In :
import math

# Given
# power dissipated at a given voltage per unit capacitance depends only on w*math.tan(delta)
# at f=60 #in Hz.
f = 60.0                # in Hz.
w = 2*math.pi*f
# let x=math.tan(delta)
x_PC = 9*10**-4         # Ploy-carbonate
x_SR = 2.25*10**-2      # Silicone rubber
x_E = 4.7*10**-2        # Epoxy with mineral filler

# Calculations and Results
p_PC = w*x_PC
p_SR = w*x_SR
p_E = w*x_E
a = min(p_PC, p_SR, p_E)
print("The minimum w*math.radians(math.tan(delta) is {0:.4f} which corresponds to polycarbonate".format(a))
print("Hence the lowest power dissipation per unit capacitance at a given voltage "
"corresponds to polycarbonate at 60Hz")

# Given
# at f=1                # in MHz.
f = 10**6               # in Hz.
w = 2*math.pi*f
# let x=math.tan(delta)
x_PC = 1*10**-2         # Ploy-carbonate
x_SR = 4*10**-3         # Silicone rubber
x_E = 3*10**-2          # Epoxy with mineral filler

# Calculations and Results
p_PC = w*x_PC
p_SR = w*x_SR
p_E = w*x_E
a = min(p_PC, p_SR, p_E)
print("The minimum w*math.radians(math.tan(delta) is {0:.4f} which corresponds to Silicone rubber".format(a))
print("Hence, the lowest power dissipation per unit capacitance at a given "
"voltage corresponds to Silicone rubber at 1MHz")

The minimum w*math.radians(math.tan(delta) is 0.3393 which corresponds to polycarbonate
Hence the lowest power dissipation per unit capacitance at a given voltage corresponds to polycarbonate at 60Hz
The minimum w*math.radians(math.tan(delta) is 25132.7412 which corresponds to Silicone rubber
Hence, the lowest power dissipation per unit capacitance at a given voltage corresponds to Silicone rubber at 1MHz


## Example 7.7 Page No : 320¶

In :
import math

# Given
# at 60 Hz
f = 60.0                    # Hz
E = 100*10**3*10**2         # in V/m
# values taken from table 7.3
epsilon_o = 8.85*10**-12    # in F/m
epsilon_r_HLPE = 2.3
epsilon_r_Alumina = 8.5
# let x=math.tan(delta)
x_HLPE = 3*10**-4
x_Alumina = 1*10**-3

# Calculations and Results
W_vol_HLPE = 2*math.pi*f*E**2*epsilon_o*epsilon_r_HLPE*x_HLPE  # in W/m3
W_vol_HLPE /= 10**3         # in mW/cm3
print("Heat dissipated per unit volume of HLPE at 60 Hz in mW/cm3 is {0:.4f}".format(W_vol_HLPE))
W_vol_Alumina = 2*math.pi*f*E**2*epsilon_o*epsilon_r_Alumina*x_Alumina
W_vol_Alumina /= 10**3      # in mW/cm3
print("Heat dissipated per unit volume of Alumina at 60 Hz in mW/cm3 is {0:.4f}".format(W_vol_Alumina))

# Given
# at 1 MHz
f = 10**6                   # Hz
x_HLPE = 4*10**-4
x_Alumina = 1*10**-3

# Calculations and Results
W_vol_HLPE = 2*math.pi*f*E**2*epsilon_o*epsilon_r_HLPE*x_HLPE  # in W/m3
W_vol_HLPE /= 10**6         # in W/cm3
print("Heat dissipated per unit volume of HLPE at 1 MHz in mW/cm3 is {0:.4f}".format(W_vol_HLPE))
W_vol_Alumina = 2*math.pi*f*E**2*epsilon_o*epsilon_r_Alumina*x_Alumina
W_vol_Alumina /= 10**6      # in W/cm3
print("Heat dissipated per unit volume of Alumina at 1 MHz in mW/cm3 is {0:.4f}".format(W_vol_Alumina))
print("The heats at 60Hz are small comparing to heats at 1MHz")

Heat dissipated per unit volume of HLPE at 60 Hz in mW/cm3 is 0.2302
Heat dissipated per unit volume of Alumina at 60 Hz in mW/cm3 is 2.8359
Heat dissipated per unit volume of HLPE at 1 MHz in mW/cm3 is 5.1158
Heat dissipated per unit volume of Alumina at 1 MHz in mW/cm3 is 47.2653
The heats at 60Hz are small comparing to heats at 1MHz


## Example 7.10 Page No : 324¶

In :
import math

# Given
# part(C)
d = 0.5                 # cm
a = d/2.0               # in cm
t = 0.5                 # in cm
Ebr_X = 217.0           # in kV/cm from table 7.5
Ebr_S = 158.0           # in kV/cm from table 7.5

# Calculations and Results
b = a+t
Vbr_X = Ebr_X*a*math.log(b/a)
print("breakdown voltage of XLPE in kV is {0:.4f}".format(Vbr_X))
Vbr_S = Ebr_S*a*math.log(b/a)
print("breakdown voltage of Silicone rubber in kV is {0:.4f}".format(Vbr_S))
# part(d)
# Given
# letE=epsiolon
Er_X = 2.3              # for XLPE
Er_S = 3.7              # for Silicone rubber
# Eair_br=Ebr
Eair_br_X = 100.0       # in kV/cm
Eair_br_S = 100.0       # in kV/cm

# Calculations and Results
# Vair_br=Eair_br*a*math.log(b/a)/Er
Vair_br_X = Eair_br_X*a*math.log(b/a)/Er_X
print("Voltage for partial discharge in a microvoid for XLPE in  kV is {0:.4f}".format(Vair_br_X))
Vair_br_S = Eair_br_S*a*math.log(b/a)/Er_S
print("Voltage for partial discharge in a microvoid for Silicone rubber in  kV is {0:.4f}".format(Vair_br_S))

breakdown voltage of XLPE in kV is 59.5997
breakdown voltage of Silicone rubber in kV is 43.3952
Voltage for partial discharge in a microvoid for XLPE in  kV is 11.9414
Voltage for partial discharge in a microvoid for Silicone rubber in  kV is 7.4231


## Example 7.11 Page No : 327¶

In :
import math

# Given
# letE=epsiolon
Er_100c = 2.69
Er_25c = 2.60
f = 1*10**3             # in Hz
w = 2*math.pi*f
C_25c = 560*10**-12     # in Farads
# Gp=w*C*math.tan(delta)
# let x=math.tan(delta)=0.002
x = 0.002

# Calculations and Results
Gp = w*C_25c*x
print("Equivalent parallel conductance at 25 degree celsius in ohm**-1 is {0:.4g}".format(Gp))

# Given
# at 100 c
x = 0.01

# Calculations and Results
C_100c = C_25c*Er_100c/Er_25c
Gp = w*C_100c*x
print("Equivalent parallel conductance at 100 degree celsius in ohm**-1 is {0:.4g}".format(Gp))

Equivalent parallel conductance at 25 degree celsius in ohm**-1 is 7.037e-09
Equivalent parallel conductance at 100 degree celsius in ohm**-1 is 3.64e-08


## Example 7.12 Page No : 331¶

In :
import math

# Given
Eo = 8.85*10**-12           # F/m2
Er = 1000.0
D = 3*10**-3                # in m
V = 5000.0                  # in V
d = 200*10**-12             # in m/V
L = 10*10**-3               # in mm

# Calculations and Results
A = math.pi*(D/2.0)**2
F = Eo*Er*A*V/(d*L)
print("Force required to spark the gap in Newton is {0:.4f}".format(F))

Force required to spark the gap in Newton is 156.3924


## Example 7.13 Page No : 333¶

In :
import math

# Given
fs = 1.0                  # in MHz
k = 0.1

# Calculations and Results
fa = fs/(math.sqrt(1-k**2))
print("fa value in MHz for given fs is {0:.4f}".format(fa))
print("thus fa-fs is only {0:.4f} kHz (which means they are very close)".format((fa-fs)*10**3))

fa value in MHz for given fs is 1.0050
thus fa-fs is only 5.0378 kHz (which means they are very close)


## Example 7.14 Page No : 334¶

In :
import math

# Given
Co = 5.0              # in pF
fa = 1.0025           # in MHz
fs = 1.0              # in MHz
R = 20.0              # in ohms

# Calculations and Results
C = Co*((fa/fs)**2-1)
print("Capacitance value in the equivalent circuit of the crystal in pF is {0:.4f}".format(C))
L = 1/(C*(2*math.pi*fs)**2)
print("Inductance value in the equivalent circuit of the crystal in Henry is {0:.4f}".format(L))

# Given
fs *= 10**6         # in Hz
C *= 10**-12        # in F

# Calculations and Results
Q = 1.0/(2*math.pi*fs*R*C)
print("Quality factor of the crystal is {0:.4g}".format(Q))

Capacitance value in the equivalent circuit of the crystal in pF is 0.0250
Inductance value in the equivalent circuit of the crystal in Henry is 1.0119
Quality factor of the crystal is 3.179e+05


## Example 7.15 Page No : 340¶

In :
# Given
P = 380*10**-6          # in C/m2/K
c = 380.0               # in J/Kg/K
# let epsilon=E
Eo = 8.85*10**-12       # in F/m
Er = 290.0
rho = 7000.0            # in Kg/m3
delta_V = 0.001         # in V
delta_t = 0.2           # in seconds

# Calculations and Results
I = (P/(rho*c*Eo*Er))**-1*delta_V/delta_t
print("Minimum radiation intensity that can be measured in W/m2 is {0:.4f}".format(I))

Minimum radiation intensity that can be measured in W/m2 is 0.0898