# 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))
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))
# 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))
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")
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")
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))
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))
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))
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))
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))
# 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))