Chapter 12 High Voltage cables

Example 12.17.1 pgno403

In [1]:
#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"Radius = %.1f cm "%r2
print"Inner radial thickness = %.1f cm "%inner
print"Outer radial thickness = %.1f cm"%outer
print"Vpeak of outer dielectric = %.2f kV"%V1peak
print"Vpeak of inner dielectric = %.1f kV"%V2peak
print"Peak voltage of cable = %.2f kV"%Vcab
print"Safe opearating voltage = %d kV"%round(rms)

#answers may vary due to round off error.

Radius = 4.8 cm 
Inner radial thickness = 2.8 cm 
Outer radial thickness = 0.2 cm
Vpeak of outer dielectric = 70.04 kV
Vpeak of inner dielectric = 4.9 kV
Peak voltage of cable = 74.94 kV
Safe opearating voltage = 53 kV

Example 12.17.2 pgno404

In [2]:
#Calculate optimum value of r

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

Radius = 2.57 cm

Example 12.17.3 pgno406

In [3]:
#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"Resistivity of insulation material = %.2f ohm.m "%round(p/10**10,2)

Resistivity of insulation material = 4.53 ohm.m

Example 12.17.4 pgno406

In [4]:
#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"C4 = %d mircoF "%c4
print"Line charging current = %.2f A "%Ic
print"Charging = %.2f kVA "%round(C,2)

C4 = 5 mircoF 
Line charging current = 18.14 A 
Charging = 314.16 kVA

Example 12.17.5 pgno408

In [5]:
#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"C2 = %.3f mircoF/Km "%C2
print"C3 = %.3f mircoF/Km "%C3
print"C4 = %.3f 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"Carging = %.2f kVAr "%round(C,1)

C2 = 0.250 mircoF/Km 
C3 = 0.650 mircoF/Km 
C4 = 0.450 mircoF/Km 
Carging = 3079.10 kVAr

Example 12.17.6 pgno409

In [6]:
#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"Conductor resistance = %.3f ohm"%rc
print"Conductor resistance for the whole leangth (Rc) = %.1f ohm"%Rc
print"Resistance of sheath (Rsh) = %.2f ohm/Km ",Rsh
print"Conductor to sheath mutual inductive reactance (Xm)= %.1f ohm/m "%Xm
print"Effective AC resistance(Ref) = %.2f ohm "%Ref
print"Reactance with sheaths open-circuit(Xc) = %.2f ohm "%Xc
print"Effective reactance per cable(Xef) = %.2f ohm "%Xef
print"Sheath loss/conductor loss = %.2f "%SlCl
print"emf induced per sheath(emf) = %.2f kV"%round(emf/1000,2)

Conductor resistance = 0.105 ohm
Conductor resistance for the whole leangth (Rc) = 8.9 ohm
Resistance of sheath (Rsh) = %.2f ohm/Km  22.8257125656
Conductor to sheath mutual inductive reactance (Xm)= 5.7 ohm/m 
Effective AC resistance(Ref) = 10.27 ohm 
Reactance with sheaths open-circuit(Xc) = 11.10 ohm 
Effective reactance per cable(Xef) = 11.04 ohm 
Sheath loss/conductor loss = 0.15 
emf induced per sheath(emf) = 2.28 kV

Example 12.17.7 pgno410

In [7]:
#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"If the sheaths are bonded at one end, the voltage between them at the other end = %f V/km"%(E*(3)**0.5)
print"\nInduced sheath voltage per Km = %.1f V/km"%round(E,1)

If the sheaths are bonded at one end, the voltage between them at the other end = 46.131881 V/km

Induced sheath voltage per Km = 26.6 V/km

Example 12.17.8 pgno411

In [8]:
#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
print V
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"Peak voltage of the conductor V = %.2f kV"%V
print"V1 = %.2f kV"%V1
print"V2 = %.2f kV"%V2
print"Maximum stress without sheaths = %.2f kV/cm"%(Emax0)
print"Minimum stress without sheaths = %.2f kV/cm"%(Emin0)
print"Maximum stress with sheaths = %.2f kV/cm"%(Emax)
print"Minimum stress with sheaths = %.2f kV/cm"%(Emin)

# Answers vary due to round off errors.

53.8887743412
Peak voltage of the conductor V = 53.89 kV
V1 = 41.10 kV
V2 = 23.90 kV
Maximum stress without sheaths = 55.30 kV/cm
Minimum stress without sheaths = 20.87 kV/cm
Maximum stress with sheaths = 38.59 kV/cm
Minimum stress with sheaths = 27.89 kV/cm

Example 12.17.9 pgno412

In [9]:
#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"V1 = %f kV/cm"%V1
print"V2 = %f kV/cm"%V2
print"\nEmax = %.2f kV/cm"%Emax

# Answers may vary due to round off error.

V1 = 45.562692 kV/cm
V2 = 23.204074 kV/cm

Emax = 47.50 kV/cm

Example 12.17.10 pgno412

In [10]:
#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"E1max = %.2f kV/cm"%round(E1max,1)
print"E2max = %.2f kV/cm"%round(E2max,2) 
# Answer vary from the text due to round off 

E1max = 51.50 kV/cm
E2max = 32.18 kV/cm