Chapter 12 High Voltage cables

Example 12_1 pgno403

In [2]:
#Chapter 12, Exmaple 1, page 403
#Calculate radial thickness of insulating layer
from math import log
#based on equation 12.15 and v1alues of E1 and E2 
E1 = 40. # kV/cm
E2 = 25. # kV/cm
ep1 = 6. # permittives of the material
ep2 = 4. #permittives of the material
d1 = 4. # cm
d2 = 10. # cm
r1 = 2. # cm
r2 = (E1*ep1*2)/(E2*ep2)
inner = r2-(d1/2)
outer = (d2/2)-r2
#based on equation 12.16
V1peak = E1*r1*log(r2/r1) # inner dielectric
V2peak = E2*r2*log(d2/(2*r2)) # outter dielectric
Vcab = V1peak+V2peak # Peak volatge of cable
rms = Vcab/(2)**0.5
print"\n Radius =  cm ",r2
print"\n Inner radial thickness = cm ",inner
print"\n Outer radial thickness =  cm",outer
print"\n Vpeak of outer dielectric =  kV", V1peak
print"\n Vpeak of inner dielectric =  kV", V2peak
print"\n Peak voltage of cable =  kV", Vcab
print"\n Safe opearating voltage =  kV", round(rms)

 Radius =  cm  4.8

 Inner radial thickness = cm  2.8

 Outer radial thickness =  cm 0.2

 Vpeak of outer dielectric =  kV 70.0374989883

 Vpeak of inner dielectric =  kV 4.89863934243

 Peak voltage of cable =  kV 74.9361383307

 Safe opearating voltage =  kV 53.0

Example 12_2 pgno404

In [4]:
#Chapter 12, Exmaple 2, page 404
#Calculate optimum value of r

#Based on equation 12.17
V1 = 100 # kV
V2 = 55 # kV
r = V1*(2)**0.5/V2
print"\n Radius =  cm ",round(r,2)

 Radius =  cm  2.57

Example 12_3 pgno406

In [9]:
#Chapter 12, Exmaple 3, page 406
#Calculate resistivity
from math import log,pi
l = 10**4 # cable length in m
Rr = 3/1.5 # R/r ratio
ins = 0.5*10**6 # insulation in ohms
p = 2*pi*l*ins/log(Rr)
print"\n Resistivity of insulation material = 10**10 ohm/m ",round(p/10**10,2)

 

 Resistivity of insulation material = 10**10 ohm/m  4.53

Example 12_4 pgno406

In [11]:
#Chapter 12, Exmaple 4, page 406
#Calculate resistivity
from math import pi
# Baased on Equation 12.1*10**2
c4 = 0.5*10**2/10 # micro F
Ic = 2*10**4*2*pi*5*50*10**-6/(3)**0.5
C = ((3)**0.5*10000*Ic)*(10**-9*10**6)
print"\n C4 =  mircoF ",c4
print"\n Line charging current =  A ",Ic
print"\n Charging =  kVA ",round(C,2)

 C4 =  mircoF  5.0

 Line charging current =  A  18.1379936423

 Charging =  kVA  314.16

Example 12_5 pgno408

In [13]:
#Chapter 12,Example 5, page 408
#Calculate capasitance and kVAr 
from math import pi
#(a) Using the notations used in FiVgs. 12.15 and 12.16
C2 = 0.75/3 # microF/km
C3 = (0.6*3-2*C2)/2 # microF/km
C4 = (C2+C3)/2 # microF/km
print"\n C2 =  mircoF/Km ",C2
print"\n C3 =  mircoF/Km ",C3
print"\n C4 = mircoF/Km ",C4
#(b)Capacitance of 10 km between 2 cores
V = 33*10**3
w = 2*pi*50
C = 2*V**2*w*C4*10*10**-9
print"\n Carging =  kVAr ",round(C,1)

 C2 =  mircoF/Km  0.25

 C3 =  mircoF/Km  0.65

 C4 = mircoF/Km  0.45

 Carging =  kVAr  3079.1

Example 12_6 pgno409

In [15]:
#Chapter 12,Example 6, page 409
#Determine the efective electrical parameters 
from math import pi,log
rc = 0.0875*(1+0.004*50) # conductor resistance in ohm/km
Rc = 0.105*85 # ohm
w = 2*pi*50
Rsh = 23.2*10**-6*85*10**5/(pi*(3**2-2.5**2)) # Resistance of sheath
D = 8.
rsh = 1./2.*(2.5+3)
Xm = w*2*log(D/rsh)*10**-7*85000
Ref = Rc + Xm**2*Rsh/(Rsh**2+Xm**2) # Effective AC resistance 
Xc = 11.1# reactance with sheaths open-circuit
Xef = Xc-(Xm**2/(Rsh**2+Xm**2)) #Effective reactance per cable
SlCl = Rsh*Xm**2/(Rc*(Rsh**2+Xm**2)) # Sheath loss/conductor loss
I = 400 # A
emf = Xm*I # emf induced per sheath
print"\n Conductor resistance =  ohm",rc
print"\n Conductor resistance for the whole leangth (Rc) =  ohm",Rc
print"\n Resistance of sheath (Rsh) =  ohm/Km ",Rsh
print"\n Conductor to sheath mutual inductive reactance (Xm)=  ohm/m ",Xm
print"\n Effective AC resistance(Ref) =  ohm ",Ref
print"\n Reactance with sheaths open-circuit(Xc) =  ohm ",Xc
print"\n Effective reactance per cable(Xef) =  ohm ",Xef
print"\n Sheath loss/conductor loss =  ",SlCl
print"\n emf induced per sheath(emf) =  kV",round(emf/1000,2)

 Conductor resistance =  ohm 0.105

 Conductor resistance for the whole leangth (Rc) =  ohm 8.925

 Resistance of sheath (Rsh) =  ohm/Km  22.8257125656

 Conductor to sheath mutual inductive reactance (Xm)=  ohm/m  5.70302447331

 Effective AC resistance(Ref) =  ohm  10.2661818113

 Reactance with sheaths open-circuit(Xc) =  ohm  11.1

 Effective reactance per cable(Xef) =  ohm  11.0412424998

 Sheath loss/conductor loss =   0.150272471857

 emf induced per sheath(emf) =  kV 2.28

Example 12_7 pgno410

In [17]:
#Chapter 12,Example 7, page 410
#Determine the induced sheath voltage 
from math import log
D = 15. # cm
rsh = 5.5/2 # Sheath diameter converted to radius in cm
I = 250. # A
E = 2*10**-7*314*I*log(D/rsh)*10**3

print"\n If the sheaths are bonded at one end, the voltage between them at the other end = =  V/km",E*(3)**0.5
print"\n Induced sheath voltage per Km =  V/km",round(E,1)

 Induced sheath voltage per Km =  V/km 26.6

 If the sheaths are bonded at one end, the voltage between them at the other end = =  V/km 46.1318808794

Example 12_9 pgno412

In [19]:
#Chapter 12,Example 9, page 412
#Determine the maximum stress 
from math import log
Emax = 47.5 # kV
b = 2.65 # cm
a = 1. # cm
ba = 0.55*3 # 1/3(b-a)
r1 = 1.55 # cm
r2 = 2.1 # cm2Vr = 2.65 # cm  
V = 53.8 # kV
alpha = ba**(1/3)
# based on the example 12_8 
#calculating VEmax1, Emax2, Emax3 
x = 1/(a*log(r1/a))
y = 1/(r1*log(r2/r1))
z = 1/(r2*log(b/r2))
VV1 = Emax/x
V1V2 = Emax/y
V2 = Emax/z
V1 = V2+(Emax/y)

print"\n V1 =  kV/cm",V1
print"\n V2 =  kV/cm",V2
print"\n Emax =  kV/cm",Emax
# Answers may vary due to round off error.

 V1 =  kV/cm 45.562691669

 V2 =  kV/cm 23.2040739531

 Emax =  kV/cm 47.5

Example 12_8 pgno411

In [26]:
#Chapter 12,Example 8, page 411
#Determine the maximum stress 
from math import log
ba = 5.3/2 # b/a
alpha = ba**(1/3)
r1 = 1.385 # cm
r2 = 1.92 # cm
r = 2.65 # cm
V = 66*(2)**0.5/(3)**0.5
V2 = 23.9#V/(1+(1/alpha)+(1/alpha**2))
V1 = 41.1#(1+1/alpha)*V2
#calculating maximim and minimum stress without sheaths
Emax0 = V/1*log(r/1)
Emin0 = V/(r*log(r))
#calculating max and min stress with the sheaths
Emax = Emax0*3/(1+(alpha)+(alpha**2))
Emin = Emax/alpha
print"\n Peak voltage of the conductor V =  kV",V
print"\n V1 =  kV",V1
print"\n V2 =  kV",V2
print"\n Maximum stress without sheaths =  kV/cm",Emax0/2
print"\n Minimum stress without sheaths =  kV/cm",Emin0*2
print"\n Maximum stress with sheaths =  kV/cm",Emax/2
print"\n Minimum stress with sheaths =  kV/cm",Emin/2

# Answers vary due to round off errors.

 Peak voltage of the conductor V =  kV 53.8887743412

 V1 =  kV 41.1

 V2 =  kV 23.9

 Maximum stress without sheaths =  kV/cm 26.258912261

 Minimum stress without sheaths =  kV/cm 41.7324619433

 Maximum stress with sheaths =  kV/cm 26.258912261

 Minimum stress with sheaths =  kV/cm 26.258912261

Example 12_10 pgno412

In [28]:
#Chapter 12,Example 10, page 412
#Determine the maximum stress 
from math import log,e
a = 1 #cm
r1 = 2 # cm
b = 2.65 # cm
er1 = 4.5
er2 = 3.6
V = 53.8 # kV
ba = 5.3/2 # b/a
alpha = 1.325
E1max = V/(log(r1)+(er1/er2)*log(alpha))
E2max = V/((r1*(er2/er1)*log(r1))+log(alpha))
print"\n E1max =  kV/cm",round(E1max,1)
print"\n E2max =  kV/cm",round(E2max,2) # answer vary from the text

# Answer vary from the text due to round off 

 E1max =  kV/cm 51.5

 E2max =  kV/cm 38.69